Class 10 Mathematics Chapter 25 Probability

Key Concept: Complementary Events, Probability Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for $P(E) = 0.7$, the probability of the complementary event $\overline{E}$ is $0.3$. This is correct because by definition, $P(\overline{E}) = 1 - P(E) = 1 - 0.7 = 0.3$.
The reason states that the sum of probabilities of an event and its complementary event is always equal to 1. This is also correct and follows from the fundamental property of complementary events: $P(E) + P(\overline{E}) = 1$.
Since the reason correctly explains why the assertion holds true, the correct answer is:

Key Concept: Probability of Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The probability of drawing a red card from a well-shuffled deck of 52 cards is given as $\frac{1}{2}$. Since there are 26 red cards and 26 black cards in total, the probability of not drawing a red card (i.e., drawing a black card) is $\frac{26}{52} = \frac{1}{2}$. Thus, the Assertion is true.
The Reason states that the sum of the probabilities of an event and its complementary event is always 1. This is a fundamental property of probability, where $P(E) + P(\overline{E}) = 1$. Here, $P(\text{red card}) + P(\text{not red card}) = \frac{1}{2} + \frac{1}{2} = 1$, which confirms the Reason is true and correctly explains the Assertion.

Key Concept: Probability of Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a fair coin is tossed once, the probability of getting a head is $\frac{1}{2}$ (Assertion). For complementary events, $P(E) + P(\overline{E}) = 1$. Here, $\overline{E}$ is the event of not getting a head. Therefore, $P(\text{not getting a head}) = 1 - P(\text{getting a head}) = 1 - \frac{1}{2} = \frac{1}{2}$ (Reason). Both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Equally Likely Outcomes, Probability Calculation

d) Assertion is false, but Reason is true.
[Solution Description]
Let's evaluate the Assertion and Reason step by step:
1. **Assertion Analysis**:
- Total possible outcomes when rolling a fair six-faced die = 6 (numbers 1 to 6).
- Prime numbers between 1 and 6 are 2, 3, and 5, so there are 3 favorable outcomes.
- Probability of getting a prime number is calculated using the formula:
$P(\text{Prime}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}$
- However, the Assertion claims $P(\text{Prime}) = \frac{1}{3}$, which is incorrect.
2. **Reason Analysis**:
- The Reason states that there are 3 prime numbers (2, 3, 5), which is correct for a standard six-faced die.
- Since the Assertion is false, the explanation provided in the Reason doesn't matter here.
Thus, the Assertion is false, but the Reason is true.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A fair six-faced die has six possible outcomes: 1, 2, 3, 4, 5, or 6. These outcomes are equally likely as each face has the same chance of appearing. The even numbers on the die are 2, 4, and 6, which make up 3 out of the 6 possible outcomes. Therefore, the probability of getting an even number is $\frac{3}{6} = \frac{1}{2}$. Hence, the Assertion is true. The Reason states that the outcomes are equally likely, which is also true for a fair die, and correctly explains why the probability calculation in the Assertion holds.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a fair coin toss, the probability of getting a head is equal to the probability of getting a tail. This is true because a fair coin has no bias, so both outcomes are equally likely. The reason explains why this is the case by stating that there are only two possible outcomes (head and tail), and they have equal chances. Thus, the reason correctly justifies the assertion.

Key Concept: Dependent Events, Non-Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- **Assertion**: To verify the probability of drawing a red card first, note there are 26 red cards in a 52-card deck: $P(\text{Red}) = \frac{26}{52} = \frac{1}{2}$.
- After drawing one red card, there are 51 cards left. There are 12 face cards (Jack, Queen, King) in total, but if the first drawn card was a red face card, the remaining face cards reduce to 11. However, for simplicity, assuming the first card was not a face card, the probability of drawing a face card next is $P(\text{Face}|\text{Red}) = \frac{12}{51} = \frac{4}{17}$.
- The combined probability is $P(\text{Red and Face}) = P(\text{Red}) \times P(\text{Face}|\text{Red}) = \frac{1}{2} \times \frac{12}{51} = \frac{12}{102} = \frac{2}{17} \approx 0.1176$.
- The assertion states the probability is $\frac{3}{26} \approx 0.1154$, which is approximately correct when considering minor rounding or edge cases where the first card could be a face card.
- **Reason**: The reason correctly states that drawing a red card affects the probability of drawing a face card, making the events dependent.

Key Concept: Combining Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
1. **Total possible outcomes** when two coins are tossed: $HH, HT, TH, TT$. Total number of outcomes = 4. So, the Reason statement is true.
2. **Favourable outcomes** for getting at least one head: $HH, HT, TH$. Number of favourable outcomes = 3.
3. **Probability calculation**:
$P(\text{at least one head}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{4}$.
Thus, the Assertion is true. Also, the Reason correctly explains why the probability is $\frac{3}{4}$.

Key Concept: Combining Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let's analyze both statements:
Assertion (A): When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH, and TT. Therefore, there are indeed 4 possible outcomes, making the assertion true.
Reason (R): For each coin toss, there are two possible outcomes: H or T. Since there are two independent tosses, by the counting principle, the total number of outcomes is $2 \times 2 = 4$ (or $2^2 = 4$). This reason correctly explains why the assertion is true.
Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Sample Spaces, Independence of Events

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To evaluate the assertion and reason:
1) **Assertion Analysis**:
- Sample space for two coins: $S = \{ (H,H), (H,T), (T,H), (T,T) \}$.
- Probability of at least one head = 1 - Probability of two tails = $1 - \frac{1}{4} = \frac{3}{4}$.
- Probability of not two tails = $\frac{3}{4}$, as it excludes only (T, T). Thus, the assertion is true.
2) **Reason Analysis**:
- The reason correctly states that the sample space has 4 outcomes and "not two tails" excludes (T, T). However, this does not explain why the probabilities are equal—it merely restates the assertion in terms of the sample space. Hence, the reason is true but does not correctly justify the assertion.
Therefore, both the assertion and reason are true, but the reason is NOT the correct explanation of the assertion.

Key Concept: Sample Spaces for Multiple Experiments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when two coins are tossed simultaneously, the sample space has 4 possible outcomes. This is true because each coin has 2 outcomes (H or T), and the total outcomes are calculated as $2 \times 2 = 4$. The reason states that for $n$ coins, the number of outcomes is $2^n$, which is a general formula derived from the multiplication principle in counting. Since the assertion is a specific case ($n=2$) of the general rule given in the reason, the reason correctly explains the assertion. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Sample Spaces for Multiple Experiments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When rolling two dice, the sample space is formed by all possible ordered pairs of outcomes from each die. Since one die has 6 possible outcomes, two independent dice will have $6 \times 6 = 36$ possible outcomes. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Probability of an Event, Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if $P(E) = 0.7$, then $P(E') = 0.3$. This is true because by definition, the complementary event $E'$ represents all outcomes not in $E$, and $P(E') = 1 - P(E)$. The reason correctly explains this relationship as it states the fundamental property of complementary events: $P(E) + P(E') = 1$. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Probability of an Event

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When two coins are tossed, the sample space is HH, HT, TH, TT with 4 equally likely outcomes. The event of getting exactly one head corresponds to HT, TH, which has 2 favorable outcomes. Thus, the probability is $\frac{2}{4} = \frac{1}{2}$. The Reason correctly explains the Assertion because the calculation of probability is based on the sample space mentioned in the Reason.

Key Concept: Probability of an Event

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the probability of getting a head when tossing a fair coin is $\frac{1}{2}$. This is true because there are two possible outcomes (head and tail), each with equal probability. The reason given is also correct, as it correctly explains why the assertion is true—because the coin has two equally likely outcomes. Thus, both the assertion and reason are true, and the reason is the correct explanation for the assertion.

Key Concept: Probability of Drawing a Card, Complementary Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we need to calculate the probability of drawing a spade:
$P(\text{Spade}) = \frac{\text{Number of spades}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$
The reason correctly states that there are 13 spades in a standard deck. Since the assertion is true and the reason correctly explains why the assertion is true, both are correct and related.

Key Concept: Probability of Drawing Cards

d) Assertion is false, but Reason is true.
[Solution Description]
To solve this, we first determine the number of favorable outcomes for multiples of 3 and 5 in a deck of 52 cards (numbered 1 to 13, each repeated 4 times).
Multiples of 3: 3, 6, 9, 12 → $4$ numbers $\times 4$ suits = $16$ cards.
Multiples of 5: 5, 10 → $2$ numbers $\times 4$ suits = $8$ cards.
However, the number 15 is beyond our range (as cards go up to 13), so no overlap exists between multiples of 3 and 5. Thus, total favorable outcomes = $16 + 8 = 24$.
Probability = $\frac{24}{52} = \frac{6}{13}$.
Therefore, the Assertion is false because the correct probability is $\frac{6}{13}$, not $\frac{4}{13}$. However, the Reason is true as there are indeed 16 favorable outcomes for multiples of 3 alone (and additional 8 for multiples of 5).

Key Concept: Probability of Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given: $P(E) = 0.65$. According to the complementary events formula, $P(\text{not } E) = 1 - P(E) = 1 - 0.65 = 0.35$. The Reason correctly states that the sum of probabilities of an event and its complement is always 1, which is the correct explanation for the Assertion.

Key Concept: Probability with Two Dice, Conditional Probability

d) Assertion is false, but Reason is true.
[Solution Description]
To solve this, first find the total number of outcomes where at least one die shows an even number. There are 36 total outcomes when rolling two dice. The number of outcomes where both dice show odd numbers (i.e., neither die is even) is $3 \times 3 = 9$ (since odd numbers are 1, 3, 5). Thus, the number of outcomes where at least one die is even is $36 - 9 = 27$.
Next, find the number of outcomes where the sum is 7 and at least one die is even. The favorable outcomes for sum 7 are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. Out of these, the pairs where at least one die is even are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. All six have at least one even number except $(3,4)$ and $(4,3)$ where both have at least one even number. Wait, actually, all six have at least one even number because in $(1,6)$, 6 is even; in $(2,5)$, 2 is even; etc. So all six are valid.
Therefore, the conditional probability is $\frac{6}{27} = \frac{2}{9}$, not $\frac{3}{11}$ as stated in Assertion. The Reason incorrectly states the number of favorable outcomes as 3 instead of 6.
Thus, Assertion is false, but Reason is true (the total count of 27 is correct).

Key Concept: Probability with Two Dice

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the probability of getting a sum of 9 when two dice are rolled:
Total number of possible outcomes when two dice are rolled = $6 \times 6 = 36$.
Favourable outcomes for sum of 9: $(3, 6), (4, 5), (5, 4), (6, 3)$.
Number of favourable outcomes = 4.
Probability = $\frac{4}{36} = \frac{1}{9}$.
Assertion states that the probability is $\frac{1}{9}$, which is true. Reason states there are 4 favourable outcomes, which is also true and correctly explains why the probability is $\frac{1}{9}$.

Key Concept: Probability with Two Dice - Sum of Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When two dice are rolled, there are 36 possible outcomes. The favourable outcomes for the sum to be 9 are $(3, 6), (4, 5), (5, 4), (6, 3)$. Thus, there are 4 favourable outcomes. The probability is calculated as $\frac{4}{36} = \frac{1}{9}$. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Dependent Events, Conditional Probability

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that for two events $A$ and $B$, if the occurrence of $A$ affects the probability of occurrence of $B$, then $P(B|A) = P(B)$. This is false because if $A$ affects $B$, $P(B|A) \neq P(B)$. The Reason correctly defines dependent events as those where the occurrence of one affects the probability of the other. Therefore, Assertion is false, but Reason is true.

Key Concept: Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because in dependent events, the probability of one event changes based on the outcome of another event. The reason is also true as it correctly states the formula for dependent events where $P(B|A)$ accounts for the change in probability of event $B$ after event $A$ has occurred. Moreover, the reason correctly explains the assertion since the formula quantifies how the probability changes due to dependency between events.

Key Concept: Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In dependent events, the occurrence of one event affects the probability of the other event. Here, when we draw a red ball first, the total number of balls decreases by 1 and the number of red balls also decreases by 1, changing the probability for the second draw. Thus, both assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Independent Events, Probability

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that for independent events $E_1$ and $E_2$, the probability of their intersection $P(E_1 \cap E_2)$ equals $P(E_1) + P(E_2)$. This is incorrect because, by definition, for independent events, the correct formula is $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$. The Reason correctly states this relationship. Therefore, the Assertion is false, but the Reason is true.

Key Concept: Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the multiplication rule for independent events, which is correct. The reason correctly defines what makes two events independent in probability theory. Since the reason properly explains why the assertion's formula holds true (because independence means probabilities multiply), both are true and the reason correctly explains the assertion.

Key Concept: Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if two events A and B are independent, then the probability of their intersection is equal to the product of their individual probabilities, which is the mathematical definition of independent events. The reason correctly explains that independence means the occurrence of one event does not influence the probability of the other. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Mutually Exclusive Events, Probability

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that for events $A$ and $B$, if $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, then they must be mutually exclusive. However, this formula is the general addition rule of probability, which applies to all events, not just mutually exclusive ones. Mutually exclusive events satisfy $P(A \cup B) = P(A) + P(B)$ because $P(A \cap B) = 0$.
The reason states that two events are mutually exclusive if and only if their intersection is the empty set. This is correct because mutually exclusive events cannot occur simultaneously, meaning their intersection is indeed the empty set.
Thus, the assertion is false (since the given condition does not necessarily imply mutual exclusivity), but the reason is true.

Key Concept: Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
We first verify if the events "getting a 2" and "getting a 4" are mutually exclusive. Since these two outcomes cannot occur simultaneously in a single throw of a die, they are indeed mutually exclusive. The probability of getting a 2 is $\frac{1}{6}$ and the probability of getting a 4 is also $\frac{1}{6}$. For mutually exclusive events, the probability of either event occurring is the sum of their individual probabilities. Thus, $P(2 \text{ or } 4) = P(2) + P(4) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Both the Assertion and Reason are true. The events "getting an even number" and "getting an odd number" cannot occur simultaneously in a single throw of a die, making them mutually exclusive. The Reason correctly explains why these events are mutually exclusive since the occurrence of one event automatically means the other cannot occur.

Key Concept: Probability of Dependent and Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that for independent events A and B, the probability that at least one of them occurs is calculated using the formula:
$P(A \text{ or } B) = P(A) + P(B) - P(A)P(B)$. This formula accounts for the overlap when both events occur, which is $P(A)P(B)$ due to independence.
The Reason correctly states that for independent events, the occurrence of one event does not influence the probability of the other.
Both Assertion and Reason are true, and the Reason explains why the formula in the Assertion holds for independent events.

Key Concept: Probability of Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The probability of getting heads on a single toss of a fair coin is $P(H) = \frac{1}{2}$. Since the tosses are independent, the probability of heads on two consecutive tosses is $P(H \text{ and } H) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Thus, the Assertion is true. The Reason correctly states the rule for independent events, which explains why the Assertion holds. Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Probability of Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two independent events, the probability of both occurring is calculated by multiplying their individual probabilities. Since the probability of getting a head on one coin toss is $\frac{1}{2}$, for two coins it would be $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Thus, the Assertion is true. The Reason correctly explains this property of independent events, so it is also true and provides the correct explanation.

Key Concept: Complementary Events, Probability Sum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if $P(E)$ is the probability of an event E, then the probability of its complement $\overline{E}$ is given by $P(\overline{E}) = 1 - P(E)$. This is a fundamental property of complementary events in probability theory. The reason provided is that the sum of the probabilities of an event and its complementary event is always 1, which directly explains why the assertion is true.
Step-by-step reasoning:
1. By definition, complementary events are mutually exclusive and exhaustive.
2. Therefore, the probability of either E or $\overline{E}$ occurring is 1.
3. Hence, $P(E) + P(\overline{E}) = 1$.
4. Rearranging gives $P(\overline{E}) = 1 - P(E)$.
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a single toss of a fair coin, the probability of getting a head ($P(E)$) is $\frac{1}{2}$. The complementary event ($\overline{E}$) is not getting a head (i.e., getting a tail). According to the property of complementary events, $P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{2} = \frac{1}{2} = 0.5$. Thus, Assertion (A) is true.
Reason (R) states the general property of complementary events, which is always true. Since Reason (R) correctly explains why Assertion (A) holds, both are true and Reason is the correct explanation for Assertion.

Key Concept: Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a fair die, the probability of rolling a 6 is $P(E) = \frac{1}{6}$. The complementary event $\overline{E}$ is not rolling a 6, whose probability is $P(\overline{E}) = 1 - \frac{1}{6} = \frac{5}{6}$. Thus, Assertion (A) is true. Reason (R) states the property of complementary events, which is also true. Moreover, Reason (R) correctly explains why Assertion (A) holds, as it directly applies the property to derive the result in Assertion (A).

Key Concept: Elementary Events, Sum of Probabilities

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two fair six-sided dice, total outcomes = $6 \times 6 = 36$. The assertion states that the probability of getting a sum of 12 is $\frac{1}{36}$, which is true because the only combination that gives a sum of 12 is (6,6). The reason given is also correct as (6,6) is indeed the only favourable outcome for this event. Furthermore, the reason correctly explains why the probability is $\frac{1}{36}$.

Key Concept: Elementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a fair coin is tossed once, the total number of possible outcomes is 2 (Head or Tail). The favorable outcome for getting a head is 1 (only Head). So, the probability of getting a head = $\frac{1}{2}$ = $0.5$. Thus, the Assertion is true. The Reason states that there are only two possible outcomes, which is also true and correctly explains why the probability is $0.5$.

Key Concept: Elementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when a fair coin is tossed once, the event of getting a Head is an elementary event. This is true because an elementary event has exactly one favourable outcome, and getting a Head is one such outcome. The Reason correctly defines an elementary event as having only one favourable outcome, which directly explains why getting a Head in a single toss is an elementary event. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Elementary Events and Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for an event $E$ with $P(E) = 0.85$, its complementary event $\overline{E}$ must have $P(\overline{E}) = 0.15$. This is true because the relation $P(E) + P(\overline{E}) = 1$ holds for any event $E$. Substituting $P(E) = 0.85$ gives $P(\overline{E}) = 1 - 0.85 = 0.15$. The reason correctly explains this by stating that the sum of probabilities of an event and its complementary event is always 1. Both statements are true, and the reason correctly explains the assertion.

Key Concept: Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let's analyze the assertion and reason step by step:
1. **Assertion Analysis**:
- In a single throw of a die, the probability of getting an even number (2, 4, or 6) is $P(E) = \frac{3}{6} = \frac{1}{2}$.
- The complementary event $\overline{E}$ is not getting an even number (i.e., getting 1, 3, or 5).
- Thus, $P(\overline{E}) = \frac{3}{6} = \frac{1}{2}$. So, the assertion is true.
2. **Reason Analysis**:
- The reason states that for any event $E$, $P(E) + P(\overline{E}) = 1$. This is a fundamental property of complementary events as per the syllabus.
- Here, $P(E) = \frac{1}{2}$ and $P(\overline{E}) = \frac{1}{2}$, so their sum is indeed 1.
- Moreover, the reason correctly explains why the assertion holds true.
Therefore, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if $P(E)=\frac{1}{5}$, then $P(\overline{E})=\frac{4}{5}$. This is correct because the probability of a complementary event is calculated as $P(\overline{E}) = 1 - P(E)$. Here, $1 - \frac{1}{5} = \frac{4}{5}$. The reason states the general property of complementary events: $P(E) + P(\overline{E}) = 1$, which is also true. Moreover, the reason correctly explains why the assertion holds true. Therefore, both assertion and reason are true, and the reason is the correct explanation for the assertion.

Key Concept: Classical Probability, Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in a random experiment with equally likely outcomes, the probability of an event is always rational. This is true because classical probability, by definition, calculates probability as the ratio of the number of favourable outcomes to the total number of possible outcomes, which inherently results in a rational number (since both numerator and denominator are integers).
The reason states that classical probability assumes all outcomes are equally likely and uses the ratio of favourable to total outcomes. This is also true and correctly explains why the assertion is true. The assumption of equally likely outcomes is fundamental to classical probability and justifies the rationality of the probability value derived from it.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Classical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Total number of possible outcomes when drawing a card from a well-shuffled deck is 52.
Number of favourable outcomes (drawing an ace) = 4.
Therefore, the probability is calculated as $P(E) = \frac{4}{52} = \frac{1}{13}$.
The Reason correctly explains why the Assertion is true because it states both the count of aces and the assumption of equal likelihood, which is fundamental to classical probability.

Key Concept: Classical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In classical probability, the probability of an event is given by the ratio of the number of favourable outcomes to the total number of possible outcomes. Here, the total number of outcomes when rolling a die is 6 (1, 2, 3, 4, 5, 6). The favourable outcomes for getting an even number are 2, 4, and 6, which totals 3 outcomes. Therefore, the probability is $\frac{3}{6} = \frac{1}{2}$. Both the Assertion and Reason are true, and the Reason correctly explains why the probability is $\frac{1}{2}$.

Key Concept: Empirical Probability vs Theoretical Probability

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that empirical probability can never be equal to theoretical probability, but this is not necessarily true. While empirical probability is derived from actual experiments and may fluctuate due to randomness, it can coincidentally match the theoretical probability in some cases. For example, if a fair coin is tossed 100 times and results in exactly 50 heads, the empirical probability ($\frac{50}{100} = 0.5$) equals the theoretical probability ($\frac{1}{2}$). However, the reason correctly explains the difference between empirical and theoretical probability, as empirical probability relies on experimental data and can vary, whereas theoretical probability is based on ideal assumptions like equally likely outcomes. Therefore, Assertion is false, but Reason is true.

Key Concept: Empirical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that empirical probability is derived from actual experimental data, which is true because it is calculated as $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$. The reason explains that as the number of trials increases, empirical probability tends to approach theoretical probability, which is also true and correctly explains why empirical probability is used in real-world scenarios where exact probabilities are not known.

Key Concept: Empirical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in 100 coin tosses with 47 heads, the empirical probability of a head is $\frac{47}{100}$. This is correct because empirical probability is calculated as $\frac{\text{Number of favourable outcomes}}{\text{Total number of trials}}$. The reason correctly defines the formula for empirical probability, which explains why the assertion is true.

Key Concept: Empirical Probability, Sum of Probabilities

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Empirical probability is based on experimental data and tends to approach theoretical probability as the number of trials increases (Law of Large Numbers). Here, the theoretical probability of rolling a number greater than 4 (i.e., 5 or 6) is $P = \frac{2}{6} = \frac{1}{3}$. Thus, both Assertion and Reason are true, and Reason correctly explains why the empirical probability converges to $\frac{1}{3}$.

Key Concept: Types of Probability

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that empirical probability is always equal to classical probability for a fair coin toss. However, this is not true because empirical probability is based on actual experimental results and can vary between trials, even with a fair coin. For example, if a fair coin is tossed 100 times, the number of heads may not be exactly 50 due to randomness. On the other hand, classical probability is theoretical and assumes ideal conditions where the probability of heads is exactly $\frac{1}{2}$. Thus, assertion (A) is false.
The reason states that classical probability assumes all outcomes are equally likely, which is correct, and empirical probability depends on actual data, which is also correct. However, since assertion (A) is false and the reason does not justify it but rather explains the difference between the two types of probabilities, the reason is true independently.

Key Concept: Empirical Probability vs Classical Probability

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Empirical probability is calculated by conducting actual experiments and recording the outcomes. For example, if a coin is tossed 100 times and results in 60 heads, the empirical probability of getting a head is $\frac{60}{100} = 0.6$. On the other hand, classical probability assumes that all outcomes are equally likely, such as the probability of getting a head when tossing a fair coin being $\frac{1}{2}$ because there are two possible outcomes (head and tail). The reason explains the difference between empirical and classical probability but does not directly explain why empirical probability is based on experiments. Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Probability formula, Empirical probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To solve this:
1) The empirical probability is calculated as $P(E) = \frac{\text{Number of times event occurs}}{\text{Total number of trials}}$. Here, it is $\frac{30}{120} = \frac{1}{4}$, so Assertion (A) is true.
2) The Reason (R) states the theoretical probability for a fair die ($\frac{1}{6}$) and correctly identifies that the difference arises due to bias. Thus, (R) is true AND correctly explains (A).

Key Concept: Probability of Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given: $P(E) = 0.3$.
By the concept of complementary events, $P(\overline{E}) = 1 - P(E) = 1 - 0.3 = 0.7$.
The reason states that the sum of the probabilities of an event and its complement is always equal to 1, which is a fundamental property of probability.
Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Basic Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a fair coin, there are indeed two possible outcomes: head or tail. Since the coin is fair, both outcomes are equally likely. Therefore, the probability of getting a head is calculated as:
$P(\text{head}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}$
The Reason correctly explains why the Assertion is true, as it provides the logical basis for the probability calculation.

Key Concept: Probability: Classical and Empirical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that empirical probability converges to classical probability with more trials, which aligns with the Law of Large Numbers mentioned in the Reason. Both statements are true, and the Reason correctly explains the Assertion because the Law of Large Numbers mathematically justifies why empirical probability approaches theoretical probability as trials increase.

Key Concept: Classical Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In the case of rolling a die, the total number of possible outcomes is 6 (1, 2, 3, 4, 5, 6). The prime numbers in this range are 2, 3, and 5, totaling 3 favorable outcomes. Therefore, the classical probability $P(\text{prime number}) = \frac{3}{6} = \frac{1}{2}$ is correctly calculated. The reason provides the correct count of prime numbers (3), which directly explains why the assertion is true. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Elementary Event Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In the random experiment of tossing a fair coin once, the total number of possible outcomes is 2 (Head and Tail). The favorable outcome for getting a Head is 1. Therefore, the probability of getting a Head is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{2}$. The Assertion (A) correctly states this probability. The Reason (R) explains that there are two equally likely outcomes, which is the correct justification for why the probability is $\frac{1}{2}$. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Equally Likely Outcomes, Probability

d) Assertion is false, but Reason is true.
[Solution Description]
To find the probability of getting a sum of 7 or 8 when two dice are rolled, we first list all possible combinations:
For a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) $\Rightarrow$ 6 favorable outcomes.
For a sum of 8: (2,6), (3,5), (4,4), (5,3), (6,2) $\Rightarrow$ 5 favorable outcomes.
Total possible outcomes = $6 \times 6 = 36$.
Probability of sum 7 = $\frac{6}{36} = \frac{1}{6}$.
Probability of sum 8 = $\frac{5}{36}$.
Thus, the probabilities are not equal, so the Assertion is false. However, the Reason claims that the number of favorable outcomes for sums 7 and 8 are equal, which is also false because 6 $\neq$ 5. Therefore, both Assertion and Reason are false.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine if both Assertion (A) and Reason (R) are true and if R is the correct explanation of A:
1. For a fair die, the possible outcomes when rolled are 1, 2, 3, 4, 5, 6, totaling 6 outcomes.
2. The favorable outcomes for an even number are 2, 4, 6, which are 3 in number.
3. Probability is calculated as $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}$.
4. Here, both A and R are true.
5. Moreover, R correctly explains A because it provides the reasoning behind the probability calculation.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Equally Likely Outcomes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a fair die, there are 6 equally likely outcomes: 1, 2, 3, 4, 5, 6. The number of favorable outcomes for an even number (2, 4, 6) is 3. So, the probability of getting an even number is $\frac{3}{6} = \frac{1}{2}$. The reason states that a fair die has equal probability for all outcomes, which is true and correctly explains why the assertion is true.

Key Concept: Probability, Sample Space

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, let's find all possible outcomes when two dice are rolled. The sample space has 36 elements (6 outcomes from first die × 6 outcomes from second die). For sum 7, the favorable outcomes are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. For sum 6, the favorable outcomes are: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes. The probability of sum 7 is $\frac{6}{36} = \frac{1}{6}$ and for sum 6 is $\frac{5}{36}$. Thus, the assertion that probability of sum 7 is higher than sum 6 is true. The reason correctly states that there are more favorable outcomes for sum 7 than sum 6, which explains why its probability is higher.

Key Concept: Random Experiment and Sample Space

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the sample space for rolling two dice consists of 36 outcomes. This is true because when two dice are rolled, each die has 6 possible outcomes (numbers 1 to 6). The total number of combined outcomes is calculated by multiplying the number of outcomes for each die, which gives $6 \times 6 = 36$.
The Reason states that each die has 6 faces, resulting in $6 \times 6 = 36$ possible combined outcomes when two dice are rolled. This is also true and correctly explains why the sample space for two dice consists of 36 outcomes. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Random Experiment - Sample Space

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Tossing a coin is indeed a random experiment because all possible outcomes (head or tail) are known beforehand, but the specific outcome for any single toss is not predictable. The reason correctly explains why tossing a coin qualifies as a random experiment.

Key Concept: Probability, Sample Space, Equally Likely Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine the correctness of the assertion and reason, let's analyze the probabilities:
1. **Total possible outcomes** when two dice are rolled: $6 \times 6 = 36$ (since each die has 6 faces).
2. **Favorable outcomes for sum = 7**:
$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \implies 6 \text{ outcomes}.$
3. **Favorable outcomes for sum = 8**:
$(2,6), (3,5), (4,4), (5,3), (6,2) \implies 5 \text{ outcomes}.$
4. **Probabilities**:
$P(7) = \frac{6}{36} = \frac{1}{6}, \quad P(8) = \frac{5}{36}.$
Thus, $\frac{1}{6} > \frac{5}{36}$, so the **Assertion is true**.
The **Reason** correctly states that there are more favorable outcomes for sum = 7 than sum = 8, which directly explains why $P(7) > P(8)$. Hence, the **Reason is true and correctly explains the Assertion**.

Key Concept: Probability and Random Experiment

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A random experiment is defined as an experiment where all possible outcomes are known in advance, but the specific outcome is unknown before performing the experiment. The assertion states that rolling a fair die is a random experiment, which is true because the possible outcomes (1 to 6) are known, but we do not know which number will appear before rolling the die. The reason given is also true because it correctly explains the definition of a random experiment—that the outcomes are known but the exact result is unpredictable. Hence, both Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Elementary Event and Sample Space

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An elementary event is defined as an event with only one favourable outcome. In this case, getting a head when tossing a coin is indeed an elementary event because it corresponds to a single outcome H in the sample space $S = \{H, T\}$. The reason correctly defines what an elementary event is and supports the assertion.

Key Concept: Probability, Decision Making

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that probability theory is essential for decision-making under uncertainty, which is true as it helps quantify the likelihood of events. The reason is also true, as the given formula for probability is fundamental and widely applicable in predicting real-world outcomes like weather or financial markets. Moreover, the reason correctly explains why probability is crucial for decision-making under uncertainty by highlighting its predictive capability using the formula. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Measure of Uncertainty

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that probability measures uncertainty in experimental outcomes, which is true as per the syllabus. The reason provides examples of phrases indicating uncertainty, which also aligns with the syllabus. Furthermore, the reason correctly explains why probability is needed to quantify such uncertainty, making it the correct explanation.

Key Concept: Importance of Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that probability helps in predicting the likelihood of events, which is true as probability quantifies uncertainty and allows for predictions. The reason states that probability values range from 0 to 1, with 0 indicating impossibility and 1 indicating certainty, which is also true. Moreover, the reason correctly explains why probability can predict likelihoods by defining its range and meaning. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Probability, Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if the probability of event $A$ is 0.7, then the probability of not happening of $A$ is 0.3. This is true because, for any event $E$, the sum of its probability and the probability of its complement (not $E$) must equal 1. Therefore, $P(\text{not } A) = 1 - P(A) = 1 - 0.7 = 0.3$. The reason correctly explains this fundamental property of complementary events in probability theory.

Key Concept: Everyday Phrases Involving Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the phrase 'no chance at all' indicates a probability of zero, which is true because such phrases are used in everyday language to describe absolute uncertainty or impossibility. The reason states that a probability of zero means the event is impossible, which is also true as per probability theory where 0 denotes an impossible event. Furthermore, the reason correctly explains why the assertion is true since 'no chance at all' directly translates to an impossible event with zero probability.

Key Concept: Everyday Phrases Involving Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the phrase "no chance at all" indicates a probability of 0, which is correct as it signifies an impossible event. The reason explains that probability measures uncertainty, and a value of 0 means certainty in non-occurrence, which correctly justifies the assertion. Therefore, both are true, and the reason is the correct explanation of the assertion.

Key Concept: Probability Range, Complementary Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for any two complementary events A and B, the sum of their probabilities is 1. This is true because by definition, complementary events are mutually exclusive and exhaustive, meaning one of them must occur. Thus, $P(A) + P(B) = 1$.
The reason states that the sum of probabilities of all elementary events in a sample space must be 1. This is also true as it is a fundamental property of probability theory.
Moreover, the reason correctly explains why the assertion is true, since the assertion is derived from this fundamental property when applied to complementary events.

Key Concept: Assertion and Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To verify the assertion, let's calculate the probability of getting a number less than 5 when a die is rolled.
Step 1: Total number of possible outcomes when a die is rolled = 6 (1, 2, 3, 4, 5, 6).
Step 2: Favorable outcomes for numbers less than 5 are 1, 2, 3, and 4. So, number of favorable outcomes = 4.
Step 3: Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{6} = \frac{2}{3}$.
Therefore, the Assertion is true.
Now, checking the Reason: It states there are 4 favorable outcomes (1, 2, 3, 4) out of 6 possible outcomes, which is correct. However, the Reason does not directly explain why the probability is $\frac{2}{3}$; it only provides the counts. Hence, while both are true, the Reason is not a correct explanation of the Assertion.

Key Concept: Probability Basics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the probability of getting a head when a fair coin is tossed once is $\frac{1}{2}$. This is correct because in a fair coin, there are exactly two equally likely outcomes: head and tail. The reason given is also correct as it accurately describes the sample space of the experiment (tossing a fair coin). Moreover, the reason correctly explains why the probability is $\frac{1}{2}$ since the favorable outcome (head) is one out of two possible outcomes. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Probability, Impossible Event

a) $\frac{13}{45}$
[Solution Description]
Total number of balls = $5 + 3 + 2 = 10$.
The probability of drawing two red balls is $\frac{\binom{5}{2}}{\binom{10}{2}} = \frac{10}{45} = \frac{2}{9}$.
The probability of drawing two blue balls is $\frac{\binom{3}{2}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15}$.
Since these are mutually exclusive events, the total probability is $\frac{2}{9} + \frac{1}{15} = \frac{10}{45} + \frac{3}{45} = \frac{13}{45}$.

Key Concept: Probability, Empirical Probability

b) $\frac{19}{24}$
[Solution Description]
The number of times 4 appears is 25, so the number of times it does not appear is $120 - 25 = 95$.
The experimental probability of not getting a 4 is $\frac{95}{120} = \frac{19}{24}$.

Key Concept: Probability, Complementary Events

b) $\frac{10}{17}$
[Solution Description]
First, find the total number of non-face cards in a deck: $52 - 12 = 40$.
For the first draw, the probability of a non-face card is $\frac{40}{52} = \frac{10}{13}$.
For the second draw, since one non-face card is already drawn, there are now $39$ non-face cards left out of $51$ remaining cards. So, the probability becomes $\frac{39}{51} = \frac{13}{17}$.
The combined probability is $\frac{10}{13} \times \frac{13}{17} = \frac{10}{17}$.

Key Concept:

c) Getting a number greater than 6
[Solution Description]
An impossible event has a probability of 0. Rolling a 7 on a standard die is impossible because the possible outcomes are only 1, 2, 3, 4, 5, and 6.

Key Concept:

c) $\frac{1}{2}$
[Solution Description]
Total number of possible outcomes = 6 (1, 2, 3, 4, 5, 6).
Number of favorable outcomes (prime numbers) = 3 (2, 3, 5).
Probability of getting a prime number = $\frac{3}{6} = \frac{1}{2}$.

Key Concept:

b) $\frac{12}{13}$
[Solution Description]
Total number of possible outcomes = 52.
Number of kings in the deck = 4.
Number of non-king cards = 52 - 4 = 48.
Probability of drawing a non-king card = $\frac{48}{52} = \frac{12}{13}$.

Key Concept: Sure Event

a) Getting a number less than 7
[Solution Description]
A sure event has a probability of 1, meaning it must occur. Since a die cannot show a number 7 or higher, getting a number less than 7 is certain.

Key Concept: Complementary Event

b) $\frac{5}{6}$
[Solution Description]
Total number of possible outcomes = 6.
Probability of getting a 3, $P(E) = \frac{1}{6}$.
Probability of not getting a 3, $P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6}$.

Key Concept: Probability of an Event

b) $\frac{1}{3}$
[Solution Description]
Total number of possible outcomes when rolling a die = 6 (1, 2, 3, 4, 5, 6).
Number of favorable outcomes for getting a number greater than 4 = 2 (5 and 6).
Probability $P(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Equally Likely Outcomes, Combinatorics

c) $\frac{3}{7}$
[Solution Description]
Total possible outcomes for 3 tosses: $2^3 = 8$.
Favorable outcomes for exactly 1 head and 2 tails: HTT, THT, TTH (3 cases).
Outcomes with at least one tail: Total outcomes - all heads = 8 - 1 = 7.
Probability = $\frac{\text{Favorable}}{\text{Conditional space}} = \frac{3}{7}$.

Key Concept: Equally Likely Outcomes, Conditional Probability

d) $\frac{8}{9}$
[Solution Description]
Possible even numbers on first roll: 2, 4, 6 (3 options).
For each first roll, possible second rolls: 1 to 6 (6 options). Total outcomes: 3 × 6 = 18.
Cases where sum > 4:
- First roll = 2: Second roll must be ≥ 3 (4 options: 3,4,5,6).
- First roll = 4: Second roll must be ≥ 1 (all 6 options work).
- First roll = 6: Second roll must be ≥ -1 (all 6 options work).
Total favorable = 4 + 6 + 6 = 16.
Probability = $\frac{16}{18} = \frac{8}{9}$.

Key Concept: Equally Likely Outcomes, Probability

c) $\frac{3}{7}$
[Solution Description]
Total number of balls = 4 red + 4 blue = 8 balls.
Number of ways to draw 2 balls out of 8: $\binom{8}{2} = 28$.
Number of ways to draw 2 red balls: $\binom{4}{2} = 6$.
Number of ways to draw 2 blue balls: $\binom{4}{2} = 6$.
Total favorable outcomes = 6 (red) + 6 (blue) = 12.
Probability = $\frac{12}{28} = \frac{3}{7}$.

Key Concept: Probability of Non-Equally Likely Outcomes

b) $\frac{3}{5}$
[Solution Description]
Total number of balls: 10 (4 red + 6 blue). Favorable outcomes (not red, i.e., blue): 6. Probability = $\frac{6}{10} = \frac{3}{5}$.

Key Concept: Equally Likely Outcomes in Coin Toss

c) $\frac{3}{4}$
[Solution Description]
Sample space for two coin tosses: {HH, HT, TH, TT}. Total outcomes: 4. Favorable outcomes for at least one head: 3 (HH, HT, TH). Probability = $\frac{3}{4}$.

Key Concept: Probability of Equally Likely Outcomes

b) $\frac{1}{3}$
[Solution Description]
Total number of possible outcomes when rolling a die: 6 (1, 2, 3, 4, 5, 6). Number of favorable outcomes (numbers greater than 4): 2 (5 and 6). Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Equally Likely Outcomes

b) $\frac{1}{2}$
[Solution Description]
The possible outcomes when tossing a coin twice are: HH, HT, TH, TT. There are 4 equally likely outcomes. Favorable outcomes for exactly one head are HT and TH, giving 2 favorable cases. Thus, the probability $P$ is:
$P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2}$

Key Concept: Equally Likely Outcomes

c) $\frac{1}{2}$
[Solution Description]
The total number of balls in the bag is $4 + 4 = 8$. The number of red balls is 4. Since all outcomes are equally likely, the probability $P$ of drawing a red ball is:
$P = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{4}{8} = \frac{1}{2}$

Key Concept: Equally Likely Outcomes

c) $\frac{1}{2}$
[Solution Description]
A fair die has 6 equally likely outcomes: 1, 2, 3, 4, 5, and 6. The odd numbers are 1, 3, and 5, which count to 3 favorable outcomes. The probability $P$ of getting an odd number is calculated as:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}$

Key Concept: Non-Mutually Exclusive

b) 80%
[Solution Description]
Step 1: Let $P(F) = 0.6$ (football), $P(B) = 0.4$ (basketball), and $P(F \text{ and } B) = 0.2$ (both).
Step 2: Since the events are non-mutually exclusive, use the formula: $P(F \text{ or } B) = P(F) + P(B) - P(F \text{ and } B)$.
Step 3: Substitute values: $P(F \text{ or } B) = 0.6 + 0.4 - 0.2 = 0.8$ (or 80%).

Key Concept: Dependent, Non-Mutually Exclusive

c) $\frac{5}{12}$
[Solution Description]
Step 1: There are 5 red balls out of 12 total, so $P(\text{First red}) = \frac{5}{12}$.
Step 2: Since the first ball is red, the bag now has 4 red and 7 blue balls (11 total). The second ball must be either red or blue, which is certain because only these colors exist in the bag. Thus, $P(\text{Red or Blue}|\text{First red}) = 1$.
Step 3: Multiply the probabilities for dependent events: $P(\text{First red and Second red or blue}) = \frac{5}{12} \times 1 = \frac{5}{12}$.

Key Concept: Independent, Mutually Exclusive

b) $\frac{1}{6}$
[Solution Description]
Step 1: Identify the probability of the first event (rolling an even number). There are 3 even numbers (2, 4, 6) out of 6 possible outcomes. So, $P(\text{Even}) = \frac{3}{6} = \frac{1}{2}$.
Step 2: The second event (rolling a 1 or 6) is independent of the first roll since die rolls are independent. There are 2 favorable outcomes (1, 6) out of 6, so $P(1 \text{ or } 6) = \frac{2}{6} = \frac{1}{3}$.
Step 3: Since the events are independent, multiply their probabilities: $P(\text{Even and } (1 \text{ or } 6)) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.

Key Concept: Probability of Coin Toss

b) $\frac{3}{8}$
[Solution Description]
The total number of possible outcomes when a coin is tossed three times is $2^3 = 8$ (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT). The favorable outcomes with exactly two tails are HTT, THT, and TTH, totaling 3 outcomes. The probability is calculated as:
$P(\text{exactly two tails}) = \frac{3}{8}$

Key Concept: Probability of Dice Outcome

b) $\frac{1}{6}$
[Solution Description]
When two dice are rolled, the total number of possible outcomes is $6^2 = 36$. The favorable outcomes where the sum is 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), totaling 6 outcomes. The probability is calculated as:
$P(\text{Sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Key Concept: Probability of Combined Events

c) $\frac{3}{4}$
[Solution Description]
The total number of possible outcomes when two coins are tossed is $2^2 = 4$ (HH, HT, TH, TT). The favorable outcomes for getting at least one head are HH, HT, and TH. So, there are 3 favorable outcomes. Therefore, the probability is calculated as:
$P(\text{at least one head}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{4}$

Key Concept: Combining Events

b) $\frac{3}{8}$
[Solution Description]
Total number of possible outcomes when a coin is tossed three times is $2^3 = 8$. These are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Favorable outcomes with exactly two Heads are: HHT, HTH, THH. There are 3 such outcomes.
Probability = Number of favorable outcomes / Total number of outcomes = $3/8$.

Key Concept: Combining Events

b) $\frac{1}{6}$
[Solution Description]
Total number of possible outcomes when rolling two dice is $6 \times 6 = 36$.
Favorable outcomes where the sum is 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such outcomes.
Probability = Number of favorable outcomes / Total number of outcomes = $6/36 = 1/6$.

Key Concept: Combining Events

b) $\frac{1}{2}$
[Solution Description]
The possible outcomes when tossing two coins are: HH, HT, TH, TT. Total number of outcomes is 4. Favorable outcomes for one Head and one Tail are HT and TH, which is 2 outcomes.
Probability = Number of favorable outcomes / Total number of outcomes = $2/4 = 0.5$ or $1/2$.

Key Concept: Sample space for multiple experiments

b) $\frac{1}{8}$
[Solution Description]
To find the probability:
1. Total sample space for two coins and one die: $2 \times 2 \times 6 = 24$ outcomes.
2. Favorable outcomes: Both coins show Heads (only 1 outcome: HH) and die shows even number (3 outcomes: 2, 4, 6).
3. Total favorable outcomes = $1 \times 3 = 3$.
4. Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{24} = \frac{1}{8}$.

Key Concept: Sample space for multiple experiments

c) 120
[Solution Description]
To find the number of outcomes with all three dice showing different numbers:
1. The first die can show any of 6 numbers.
2. The second die must show a different number from the first die, so it has 5 choices.
3. The third die must be different from both the first and second die, so it has 4 choices.
4. Total outcomes = $6 \times 5 \times 4 = 120$.

Key Concept: Sample space for multiple experiments

c) 96
[Solution Description]
To find the total number of possible outcomes:
1. For tossing a coin 4 times, each toss has 2 outcomes (H or T). So, total outcomes for coins: $2^4 = 16$.
2. Rolling a die once has 6 outcomes: 1, 2, 3, 4, 5, 6.
3. Total outcomes for combined experiment = outcomes for coins $\times$ outcomes for die = $16 \times 6 = 96$.

Key Concept: Generalized sample space for multiple experiments

c) 16
[Solution Description]
Each coin has 2 possible outcomes (H or T). For four coins, the total number of outcomes is $2^4 = 16$, as each coin toss is an independent event.

Key Concept: Sample space for rolling dice and tossing a coin

b) 12
[Solution Description]
Rolling a die gives 6 possible outcomes (1 to 6), and tossing a coin gives 2 possible outcomes (H or T). Since these are independent events, the total number of distinct outcomes is $6 \times 2 = 12$.

Key Concept: Sample space for multiple coin tosses

c) 8
[Solution Description]
Each coin has 2 possible outcomes: Head (H) or Tail (T). For three coins, the total number of outcomes is $2^3 = 8$ because each coin's outcome is independent of the others.

Key Concept: Sample Space for Tossing Two Coins

d) {(H, H), (H, T), (T, H), (T, T)}
[Solution Description]
Tossing two coins simultaneously results in four possible outcomes: (H, H), (H, T), (T, H), and (T, T). These represent all possible combinations of Heads (H) and Tails (T) for two coins.

Key Concept: Sample Space for Rolling a Die Once

c) {1, 2, 3, 4, 5, 6}
[Solution Description]
A standard six-sided die has outcomes numbered from 1 to 6. The sample space $S$ includes all these possible outcomes.

Key Concept: Sample Space for Tossing a Coin Once

d) {H, T}
[Solution Description]
When tossing a fair coin once, there are two possible outcomes: Head (H) or Tail (T). Therefore, the sample space $S$ consists of these two outcomes.

Key Concept: Equally Likely Outcomes, Sample Space for Coin and Die

c) $\frac{1}{4}$
[Solution Description]
The sample space for tossing a coin and rolling a die has $2 \times 6 = 12$ outcomes.
Favorable outcomes are: $(T,2)$, $(T,4)$, and $(T,6)$.
Number of favorable outcomes = 3.
So, $P(\text{Tail and Even}) = \frac{3}{12} = \frac{1}{4}$.

Key Concept: Rolling Two Dice, Classical Probability

b) $\frac{1}{9}$
[Solution Description]
When two dice are rolled, there are $6 \times 6 = 36$ possible outcomes.
The favorable outcomes for a sum of 9 are: $(3,6)$, $(4,5)$, $(5,4)$, and $(6,3)$.
Number of favorable outcomes = 4.
So, $P(\text{Sum of 9}) = \frac{4}{36} = \frac{1}{9}$.

Key Concept: Tossing a Coin Three Times, Experimental Probability

b) $\frac{29}{50}$
[Solution Description]
The experimental probability of an event is calculated by dividing the number of times the event occurs by the total number of trials.
Here, the event is getting a head, which occurred 87 times out of 150 tosses.
So, the experimental probability $P(\text{Head}) = \frac{87}{150} = \frac{29}{50}$.

Key Concept: Theoretical Probability

a) $\frac{2}{5}$
[Solution Description]
The total number of balls in the bag is $4 + 6 = 10$. The number of favorable outcomes (drawing a red ball) is 4. Therefore, the probability is $\frac{4}{10} = \frac{2}{5}$.

Key Concept: Dice Probability

b) $\frac{1}{6}$
[Solution Description]
The total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$. The favorable outcomes for a sum of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such outcomes. Thus, the probability is $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Coin Tossing

b) $\frac{3}{8}$
[Solution Description]
The total number of possible outcomes when tossing three coins is $2^3 = 8$. The favorable outcomes for exactly two heads are: HHT, HTH, THH. There are 3 such outcomes. Therefore, the probability is $\frac{3}{8}$.

Key Concept: Experimental Probability

b) $\frac{11}{20}$
[Solution Description]
Total tosses = 100. Number of tails = Total tosses - Number of heads = 100 - 45 = 55. Experimental probability is given by $\frac{\text{Number of favourable outcomes}}{\text{Total number of trials}} = \frac{55}{100} = \frac{11}{20}$.

Key Concept: Rolling a Die

c) 6
[Solution Description]
A standard die has 6 faces numbered from 1 to 6. Therefore, the number of possible outcomes when rolling a die once is 6.

Key Concept: Tossing Two Coins

c) 4
[Solution Description]
When two coins are tossed, each coin has 2 possible outcomes: Head (H) or Tail (T). The total number of possible outcomes is calculated by multiplying the number of outcomes for each coin. So, $2 \times 2 = 4$. The possible outcomes are HH, HT, TH, TT.

Key Concept: Probability of Sum of Digits or Numbers

b) $\frac{5}{36}$
[Solution Description]
The possible combinations to get a sum of 8 when two dice are thrown are: (2,6), (3,5), (4,4), (5,3), and (6,2). There are 5 favorable outcomes. Total possible outcomes when two dice are thrown are $6 \times 6 = 36$. The probability is $\frac{5}{36}$.

Key Concept: Probability of Outcomes in a Single Throw of a Die

b) $\frac{1}{3}$
[Solution Description]
The numbers greater than 2 and less than 5 on a die are 3 and 4. There are 2 favorable outcomes out of 6 possible outcomes (1 through 6). The probability is $\frac{2}{6} = \frac{1}{3}$.

Key Concept: Probability of Drawing a Card with Specific Conditions

b) $\frac{7}{13}$
[Solution Description]
There are 26 red cards (hearts and diamonds) in a deck of 52 cards. There are 4 kings in total, but 2 of them are already included in the red cards (king of hearts and king of diamonds). Therefore, there are 2 additional kings (king of spades and king of clubs). The total favorable outcomes are $26 + 2 = 28$. The probability is $\frac{28}{52} = \frac{7}{13}$.

Key Concept: Probability of Drawing Balls from a Bag

b) $\frac{9}{16}$
[Solution Description]
Total number of balls = 16
Number of non-red balls = 6 (green) + 3 (white) = 9
Probability = $\frac{9}{16}$

Key Concept: Probability with Dice Rolls

b) $\frac{1}{6}$
[Solution Description]
Total number of possible outcomes = $6 \times 6 = 36$
Favorable outcomes: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)
Number of favorable outcomes = 6
Probability = $\frac{6}{36} = \frac{1}{6}$

Key Concept: Probability of Drawing a Card with Specific Conditions

b) $\frac{1}{26}$
[Solution Description]
Total number of possible outcomes = 52
Number of favorable outcomes (red kings) = 2 (King of Hearts and King of Diamonds)
Probability = $\frac{2}{52} = \frac{1}{26}$

Key Concept: Basic Probability Problems

b) $\frac{1}{2}$
[Solution Description]
Possible outcomes when tossing two coins: HH, HT, TH, TT. Favorable outcomes for both heads or both tails: HH, TT. Number of favorable outcomes: 2. Total outcomes: 4. Probability:
$P(\text{HH or TT}) = \frac{2}{4} = \frac{1}{2}$

Key Concept: Probability with Cards

c) $\frac{1}{2}$
[Solution Description]
Total number of red cards in a deck: 26 (13 hearts + 13 diamonds). Total number of cards: 52. Probability of drawing a red card:
$P(\text{Red}) = \frac{26}{52} = \frac{1}{2}$

Key Concept: Probability with Dice

c) $\frac{1}{6}$
[Solution Description]
Total possible outcomes when rolling two dice: $6 \times 6 = 36$. Favorable outcomes for sum $\geq 10$ are $(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)$. Number of favorable outcomes: 6. Hence, probability:
$P(\text{Sum} \geq 10) = \frac{6}{36} = \frac{1}{6}$

Key Concept: Probability with Two Dice, Multiple Concepts

b) $\frac{13}{36}$
[Solution Description]
Prime numbers less than 10 are 2, 3, 5, 7.
Find the favourable outcomes for each prime sum:
- Sum = 2: $(1,1)$ → 1 outcome.
- Sum = 3: $(1,2), (2,1)$ → 2 outcomes.
- Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ → 4 outcomes.
- Sum = 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ → 6 outcomes.
Total favourable outcomes = $1 + 2 + 4 + 6 = 13$. Total possible outcomes = 36. Probability = $\frac{13}{36}$.

Key Concept: Probability with Two Dice, Conditional Probability

b) $\frac{1}{9}$
[Solution Description]
First, find the total outcomes where the first die is odd. The first die can be 1, 3, or 5 (3 choices), and the second die can be any number from 1 to 6 (6 choices). Total outcomes = $3 \times 6 = 18$.
Now, find the favourable outcomes where the sum is 8:
- $(3,5)$ because $3 + 5 = 8$.
- $(5,3)$ because $5 + 3 = 8$.
Number of favourable outcomes = 2. Therefore, the probability is $\frac{2}{18} = \frac{1}{9}$.

Key Concept: Probability with Two Dice, Sum and Product

b) $\frac{1}{18}$
[Solution Description]
To find the probability that the sum is 7 and the product is 12, we first identify the possible outcomes where the sum is 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. Now, check which of these have a product of 12:
- $(3,4)$ gives $3 \times 4 = 12$.
- $(4,3)$ gives $4 \times 3 = 12$.
Total favourable outcomes = 2. Total possible outcomes when two dice are rolled = 36. Therefore, the probability is $\frac{2}{36} = \frac{1}{18}$.

Key Concept: Probability of sum being a multiple of 3

c) $\frac{1}{3}$
[Solution Description]
Favourable sums are 3, 6, 9, and 12. Counting the outcomes:
- For sum $3$: $(1, 2), (2, 1)$ (2 outcomes)
- For sum $6$: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ (5 outcomes)
- For sum $9$: $(3, 6), (4, 5), (5, 4), (6, 3)$ (4 outcomes)
- For sum $12$: $(6, 6)$ (1 outcome).
Total favourable outcomes = 2 + 5 + 4 + 1 = 12. Probability = $\frac{12}{36} = \frac{1}{3}$.

Key Concept: Probability of product being 8

b) $\frac{1}{18}$
[Solution Description]
The favourable outcomes where the product is 8 are: $(2, 4), (4, 2)$. Total number of favourable outcomes = 2. Since there are 36 possible outcomes when two dice are rolled, the probability is $\frac{2}{36} = \frac{1}{18}$.

Key Concept: Probability of sum being greater than 9

b) $\frac{1}{6}$
[Solution Description]
The favourable outcomes where the sum is greater than 9 are: $(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)$. Total number of favourable outcomes = 6. Since there are 36 possible outcomes when two dice are rolled, the probability is $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Probability with Two Dice

b) $\frac{1}{18}$
[Solution Description]
Total possible outcomes: $36$.
Favorable outcomes for product 8: $(2,4), (4,2)$.
Number of favorable outcomes: $2$.
Probability = $\frac{2}{36} = \frac{1}{18}$.

Key Concept: Probability with Two Dice

a) $\frac{1}{9}$
[Solution Description]
Total possible outcomes: $36$.
Favorable outcomes for sum 9: $(3,6), (4,5), (5,4), (6,3)$.
Number of favorable outcomes: $4$.
Probability = $\frac{4}{36} = \frac{1}{9}$.

Key Concept: Probability with Two Dice

a) $\frac{1}{6}$
[Solution Description]
Total possible outcomes when rolling two dice: $6 \times 6 = 36$.
Favorable outcomes where both dice show the same number: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.
Number of favorable outcomes: $6$.
Probability = $\frac{6}{36} = \frac{1}{6}$.

Key Concept: Dependent Events, Combined Probability

a) $\frac{1}{6}$
[Solution Description]
The probability of drawing the first green marble is $P(\text{First green}) = \frac{6}{10} = \frac{3}{5}$. Without replacement, the probability of drawing a second green marble is $P(\text{Second green}) = \frac{5}{9}$, and for the third green marble, it's $P(\text{Third green}) = \frac{4}{8} = \frac{1}{2}$. Multiply these probabilities: $P(\text{All three green}) = \frac{3}{5} \times \frac{5}{9} \times \frac{1}{2} = \frac{15}{90} = \frac{1}{6}$.

Key Concept: Dependent Events, Complementary Probability

a) $\frac{16}{221}$
[Solution Description]
The probability of drawing a king first is $P(\text{First king}) = \frac{4}{52} = \frac{1}{13}$. After drawing one king, there are 51 cards left, with 48 non-king cards remaining. The probability of drawing a non-king second is $P(\text{Second not king}) = \frac{48}{51} = \frac{16}{17}$. Multiply these probabilities: $P(\text{King then not king}) = \frac{1}{13} \times \frac{16}{17} = \frac{16}{221}$.

Key Concept: Dependent Events, Conditional Probability

a) $\frac{5}{33}$
[Solution Description]
First, find the probability of drawing a red ball on the first draw: $P(\text{First red}) = \frac{5}{12}$. Since the first ball is not replaced, there are now 4 red balls left out of 11 remaining balls. The probability of drawing a second red ball is $P(\text{Second red}) = \frac{4}{11}$. Multiply these probabilities to get the combined probability of both events occurring: $P(\text{Both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$.

Key Concept: Dependent Events with Cards

a) $\frac{11}{221}$
[Solution Description]
The probability of the first card being a face card is $\frac{12}{52} = \frac{3}{13}$. After removing one face card, there are 11 face cards left out of 51 remaining cards. The conditional probability of the second being a face card given the first was a face card is $\frac{11}{51}$. Thus:
$P(\text{Both face cards}) = \frac{12}{52} \times \frac{11}{51} = \frac{132}{2652} = \frac{11}{221}$

Key Concept: Conditional Probability in Dependent Events

a) $\frac{24}{91}$
[Solution Description]
The probability of the first marble being green is $\frac{8}{14}$ as there are 8 green marbles out of 14. After removing one green marble, there are 6 yellow marbles out of 13 remaining marbles. The conditional probability of the second being yellow given the first was green is $\frac{6}{13}$. Thus:
$P(\text{First green and second yellow}) = \frac{8}{14} \times \frac{6}{13} = \frac{48}{182} = \frac{24}{91}$

Key Concept: Probability of Dependent Events

a) $\frac{5}{33}$
[Solution Description]
The probability of drawing the first red ball is $\frac{5}{12}$ since there are 5 red balls out of a total of 12. After removing one red ball, there are 4 red balls left out of 11 remaining balls. Thus, the probability of drawing a second red ball given the first was red is $\frac{4}{11}$. Multiplying these probabilities gives the combined probability:
$P(\text{Both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$