Class 12 Mathematics Chapter 18 Probability (Laws of Probability)

Key Concept: Range of Conditional Probability

b) 0 to 1
[Solution Description]
The conditional probability $P(A|B)$ is defined as $\frac{P(A \cap B)}{P(B)}$. Since probabilities are always between 0 and 1, and $P(A \cap B) \leq P(B)$, the ratio must satisfy $0 \leq P(A|B) \leq 1$.

Key Concept: Mutually Exclusive vs Independent

c) $P(A \cap B) = P(A) \times P(B)$
[Solution Description]
For two independent events $A$ and $B$ that are not mutually exclusive:
- Their probabilities multiply to give the joint probability: $P(A \cap B) = P(A) \times P(B)$.
- They can occur simultaneously, so $P(A \cap B) \neq 0$.
- If they were mutually exclusive, $P(A \cap B) = 0$, which cannot happen for independent events with non-zero probabilities.

Key Concept: Dependent Events

c) 0.12
[Solution Description]
Let $A$ be passing Math, $B$ be passing Physics, and $C$ be passing Chemistry.
Given: $P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.4$.
The conditional probability $P(C|A \cap B) = 0.3$.
First, find $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.6} = \frac{2}{3}$.
Now, using the multiplication theorem:
$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B) = 0.6 \cdot \frac{2}{3} \cdot 0.3 = 0.12$.

Key Concept: Independent Events

c) 0.4
[Solution Description]
For independent events, the probability of both occurring is:
$P(A \cap B) = P(A) \cdot P(B)$
Substituting the given values:
$P(A \cap B) = 0.5 \times 0.8 = 0.4$

Key Concept: Useful Results, Derived Laws

a) $\frac{781}{1024}$
[Solution Description]
The probability of getting any single question wrong is $\frac{3}{4}$.
The probability of getting all 5 questions wrong is:
$\left(\frac{3}{4}\right)^5 = \frac{243}{1024}.$
The probability of getting at least one correct is:
$1 - \left(\frac{3}{4}\right)^5 = 1 - \frac{243}{1024} = \frac{781}{1024}.$

Key Concept: Mutually Exclusive Events, Addition Law for Probability

a) 0.5
[Solution Description]
The events of winning a car, bike, or TV are mutually exclusive because the contestant can win only one prize. Using the addition law for mutually exclusive events:
$P(\text{car or TV}) = P(\text{car}) + P(\text{TV}) = 0.1 + 0.4 = 0.5$

Key Concept: Multiplication Theorem for Three Events

b) $\frac{1}{22}$
[Solution Description]
Total balls initially = 5 + 4 + 3 = 12.
Probability of first ball being red: $P(A) = \frac{5}{12}$.
After drawing one red ball, remaining balls = 11. Probability of second ball being blue: $P(B|A) = \frac{4}{11}$.
After drawing red and blue, remaining balls = 10. Probability of third ball being green: $P(C|A \cap B) = \frac{3}{10}$.
Using multiplication theorem:
$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B) = \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$.

Key Concept: Independence of Complements

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that if events $A$ and $B$ are independent, then the probability of their complements' intersection equals the product of their individual complement probabilities. This is true as per the theorem on independence of complements. The reason correctly explains the assertion because it highlights the property that the independence of events extends to their complements.

Key Concept: Mutually Exclusive Events

d) Both (a) and (b)
[Solution Description]
Mutually exclusive events cannot occur simultaneously. For a die roll:
- Rolling an odd number and rolling an even number are mutually exclusive because they cannot happen at the same time.
- Rolling a 1 and rolling a 2 are mutually exclusive as only one outcome is possible in a single roll.
However, rolling a number greater than 3 and rolling an even number are not mutually exclusive since outcomes like 4 or 6 satisfy both conditions.

Key Concept: Mutually Exclusive Events

b) 0
[Solution Description]
If A and B are mutually exclusive, then $P(A \cap B) = 0$. Therefore:
$P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{P(B)} = 0.$

Key Concept: Multiplication Theorem for Three Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that for three independent events $A$, $B$, and $C$, the probability of all three occurring together is the product of their individual probabilities. This is a direct consequence of the definition of independent events.
The Reason explains that independent events are those where the occurrence of one event does not influence the probability of the others. This is the definition of independent events and directly justifies the Assertion.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Mutually Exclusive Events, Probability Rules

b) B and C only
[Solution Description]
Analyze each pair of events:
1) A (red card) and B (face card): Can both occur (e.g., red face cards like King of Hearts)
2) B (face card) and C (ace): In standard deck, aces are not face cards → mutually exclusive
3) A (red card) and C (ace): Can both occur (red aces exist)
4) All three events: Not relevant as question asks for pairs
Only pair B and C cannot occur simultaneously.

Key Concept: Multiplication Theorem

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 of their intersection ($P(A \cap B)$) equals the product of their individual probabilities ($P(A) \cdot P(B)$). This is correct because independence implies no influence of one event on the other. The Reason provides the correct explanation for the Assertion by restating the definition of independent events in terms of probabilities. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Independent Events Probability

a) 0.42
[Solution Description]
The probability that both independent events occur is the product of their individual probabilities.
Step 1: Multiply the probabilities of the two events:
$P = p_1 \times p_2 = 0.6 \times 0.7 = 0.42$
Therefore, the probability that both events occur is 0.42.

Key Concept: Conditional Probability, Simple Events

d) $\frac{1}{3}$
[Solution Description]
Total face cards = 12 (4 Jacks, 4 Queens, 4 Kings).
Favorable outcome = Drawing a Queen → 4 outcomes.
Probability = $\frac{\text{Favorable}}{\text{Total face cards}} = \frac{4}{12} = \frac{1}{3}$.

Key Concept: Multiplication Theorem for Three Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion correctly states the multiplication theorem for three dependent events, where the joint probability depends on conditional probabilities. The reason explains why the formula is structured this way—to account for dependencies between events. Since both statements are true and the reason justifies the assertion, the correct option is (a).

Key Concept: Mutually Exclusive Events

d) $\frac{1}{3}$
[Solution Description]
The events "getting a 1" and "getting a 3" are mutually exclusive because they cannot occur at the same time. The probability of each event is $\frac{1}{6}$ since the die has six faces. Using the addition rule for mutually exclusive events:
$P(1 \text{ or } 3) = P(1) + P(3) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Dependent Events

a) $\frac{2}{15}$
[Solution Description]
Since the balls are drawn without replacement, the events are dependent. The probability of drawing a black ball in the first draw is $\frac{4}{10}$. After removing one black ball, the probability of drawing another black ball becomes $\frac{3}{9}$. Thus, the combined probability is:
$P(\text{Both Black}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Key Concept: Dependent Events - Cards

b) $\frac{1}{221}$
[Solution Description]
Let event $A$: First card drawn is a queen, and event $B$: Second card drawn is a queen.
Number of queens = 4, total cards = 52.
$P(A) = \frac{4}{52} = \frac{1}{13}$.
After drawing one queen, remaining queens = 3, total cards = 51.
$P(B|A) = \frac{3}{51} = \frac{1}{17}$.
The required probability is:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221}$

Key Concept: Corollary for Repeated Events

c) 0.189
[Solution Description]
The probability of the event occurring exactly twice in three trials can be calculated using the binomial probability formula:
$P = \binom{3}{2} \times p^2 \times (1 - p)^{(3-2)} = 3 \times (0.3)^2 \times (0.7)^1 = 3 \times 0.09 \times 0.7 = 0.189$.

Key Concept: Probability of Independent Events

a) 0.21
[Solution Description]
The probability that all three events fail is given by the product of the probabilities of each event failing individually.
Step 1: Find the probability of each event failing:
$1 - p_1 = 1 - 0.3 = 0.7$
$1 - p_2 = 1 - 0.4 = 0.6$
$1 - p_3 = 1 - 0.5 = 0.5$
Step 2: Multiply these probabilities together:
$P = 0.7 \times 0.6 \times 0.5 = 0.21$
Therefore, the probability that all three events fail is 0.21.

Key Concept: Probability of 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 of their intersection is the product of their individual probabilities, which is correct as per the definition of independent events. The reason correctly explains why this is true: independence means the occurrence of one event does not influence the probability of the other. Therefore, both the assertion and reason are true, and the reason is the correct explanation for the assertion.

Key Concept: Multiplication Theorem for Three Dependent Events

b) $\frac{1}{21}$
[Solution Description]
Here, the events are dependent because the marbles are drawn without replacement. The probabilities change after each draw.
Step 1: Probability of first marble being red is $\frac{4}{9}$.
Step 2: After drawing one red marble, total marbles left are 8. Probability of second marble being green is $\frac{3}{8}$.
Step 3: Now, since one red and one green marble have been drawn, total marbles left are 7. Probability of third marble being blue is $\frac{2}{7}$.
Combined probability:
$P(R \cap G \cap B) = \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7} = \frac{24}{504} = \frac{1}{21}$

Key Concept: Simple and Compound Events

c) $\frac{2}{3}$
[Solution Description]
Rolling a die gives outcomes from \{1, 2, 3, 4, 5, 6\}. Getting a '1' is a simple event with $P(1) = \frac{1}{6}$. Getting an even number (\{2, 4, 6\}) is a compound event. Since '1' and even numbers are mutually exclusive (they cannot occur together), we use the addition rule for exclusive events: $P(1 \cup \text{even}) = P(1) + P(\text{even})$.
$P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.
Hence, total probability is $\frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}$.

Key Concept: Dependent Events

b) $\frac{2}{15}$
[Solution Description]
The probability of drawing a red ball on the first draw is $\frac{4}{10}$. Since the ball is not replaced, the probability of drawing another red ball on the second draw is $\frac{3}{9}$. Thus, the combined probability is $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.

Key Concept: Addition Rule for Exclusive Events

b) $\frac{7}{12}$
[Solution Description] Since $A$ and $B$ are mutually exclusive, use the addition rule: $P(A \cup B) = P(A) + P(B)$. Substituting the given values, $P(A \cup B) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}$.

Key Concept: Dependent Events, Multiplication Theorem

b) $\frac{2}{15}$
[Solution Description]
Total bulbs = 4 defective + 6 non-defective = 10.
Probability first bulb is defective = $\frac{4}{10} = \frac{2}{5}$.
After drawing one defective bulb, remaining defective bulbs = 3, total remaining bulbs = 9.
Probability second bulb is defective given first was defective = $\frac{3}{9} = \frac{1}{3}$.
Combined probability = $\frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.

Key Concept: Multiplication Theorem, Total Probability

c) $\frac{1}{3}$
[Solution Description]
Let $A$ be the event first ball is blue, $B$ be the event second ball is red.
We need to find $P(B/A)$.
After drawing one blue ball, remaining balls are 3 red and 6 blue.
Total remaining balls = 9.
So, $P(B/A) = \frac{\text{Number of red balls left}}{\text{Total balls left}} = \frac{3}{9} = \frac{1}{3}$.

Key Concept: Mutually Exclusive Events

a) Getting a 1 or a 2
[Solution Description]
Mutually exclusive events cannot occur simultaneously.
- Getting a 1 or 2 in a single roll: mutually exclusive.
- Getting an odd number (1, 3, 5) or even number (2, 4, 6): mutually exclusive.
- Getting a 4 or a number less than 3: not mutually exclusive (e.g., 1 is less than 3 and possible).
- Getting a prime number (2, 3, 5) or an even number (2, 4, 6): not mutually exclusive (2 is both prime and even).
The correct answer is getting a 1 or 2.

Key Concept: Conditional Probability, Dependent Events

b) $\frac{4}{51}$
[Solution Description]
Initially, there are 4 kings in 52 cards.
After drawing one queen (not a king), remaining kings = 4, remaining cards = 51.
Probability second card is king given first was queen = $\frac{4}{51}$.

Key Concept: Addition Law of Probability, Multiple Exclusive Events

c) $\frac{13}{45}$
[Solution Description]
First, find the total number of ways to draw 2 balls from 10 balls:
$\binom{10}{2} = 45$.
Next, find the number of ways to draw 2 red balls:
$\binom{5}{2} = 10$.
Then, find the number of ways to draw 2 blue balls:
$\binom{3}{2} = 3$.
Since the events are mutually exclusive, add the probabilities:
$P(\text{both red or both blue}) = \frac{10}{45} + \frac{3}{45} = \frac{13}{45}$.

Key Concept: Simple Event Definition

a) Getting a head when tossing a coin.
[Solution Description]
A simple event is a single outcome of an experiment. Rolling a die and getting a '4' is a single outcome, hence a simple event.

Key Concept: Mutually Exclusive Events, Probability Addition Rule

d) $\frac{2}{3}$
[Solution Description]
The events "rolling a number less than 3" and "rolling a number greater than 4" are mutually exclusive since they cannot occur simultaneously.
Step 1: Calculate $P(\text{less than } 3)$
Numbers less than 3 on a die are \{1, 2\}.
So, $P(\text{less than } 3) = \frac{2}{6} = \frac{1}{3}$.
Step 2: Calculate $P(\text{greater than } 4)$
Numbers greater than 4 on a die are \{5, 6\}.
So, $P(\text{greater than } 4) = \frac{2}{6} = \frac{1}{3}$.
Step 3: Apply the addition rule for mutually exclusive events:
$P(\text{less than 3 or greater than 4}) = P(\text{less than 3}) + P(\text{greater than 4}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.

Key Concept: Independent vs. Mutually Exclusive 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 are mutually exclusive, they cannot be independent. This is true because for mutually exclusive events with non-zero probabilities, $P(A \cap B) = 0$, whereas independence requires $P(A \cap B) = P(A) \cdot P(B) \neq 0$. Thus, the reason correctly explains why the assertion is true.

Key Concept: Independent and Dependent Events

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that drawing two red cards without replacement are independent events, which is false because the probability of the second draw depends on the outcome of the first draw (since the deck changes). The reason states that independence implies the probability of the second draw is unaffected by the first, which is true for independent events but does not apply here as the events are dependent. Thus, Assertion is false, but Reason is true.

Key Concept: Independent Events

b) 0.7
[Solution Description]
For independent events, the probability that at least one occurs is given by:
$P(A \cup B) = P(A) + P(B) - P(A) \times P(B)$
Substituting the given values:
$P(A \cup B) = 0.4 + 0.5 - (0.4 \times 0.5) = 0.9 - 0.2 = 0.7$

Key Concept: Probability of occurrence of exactly one of two events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the probability of exactly one of two events A and B occurring, we consider two mutually exclusive cases:
1. A occurs and B does not occur: $P(A \cap B') = P(A)P(B') = p_1 q_2$
2. B occurs and A does not occur: $P(A' \cap B) = P(B)P(A') = p_2 q_1$
Adding these probabilities gives the total probability of exactly one event occurring: $p_1 q_2 + p_2 q_1$.
Both Assertion (A) and Reason (R) are true. Additionally, Reason (R) correctly explains how the formula in Assertion (A) is derived for independent events.

Key Concept: Independent Events

b) $\frac{9}{25}$
[Solution Description]
Since the draws are independent and with replacement, the probability of drawing a red ball each time is $\frac{3}{5}$. Therefore, the combined probability is:
$P(\text{Red and Red}) = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

Key Concept: Addition Law of Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the probability of either Ravi or Shekhar winning is 0.5 given their individual probabilities as 0.2 and 0.3 respectively. This can be calculated using the Addition Law of Probability for mutually exclusive events:
$P(A \cup B) = P(A) + P(B)$.
Substituting the values:
$P(\text{Ravi} \cup \text{Shekhar}) = 0.2 + 0.3 = 0.5$.
The reason states that the events are mutually exclusive, which is true because only one participant can win the race at a time. Since the reason correctly explains why the addition law applies, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Conditional Probability - Multiplication Theorem

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the definition of independent events, if $A$ and $B$ are independent, then the occurrence of one does not affect the probability of the other. Mathematically, this means $P(A \cap B) = P(A) \cdot P(B)$.
The Reason states that for independent events, $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) \cdot P(B)}{P(B)} = P(A)$. Similarly, $P(B/A) = P(B)$. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Conditional Probability, Independent Events

c) $\frac{5}{8}$
[Solution Description]
Let:
- $P(M1) = 0.6$, $P(M2) = 0.4$,
- $P(D|M1) = 0.02$, $P(D|M2) = 0.05$.
First, find the total probability of a defective bulb:
$P(D) = P(M1) \cdot P(D|M1) + P(M2) \cdot P(D|M2) = 0.6 \times 0.02 + 0.4 \times 0.05 = 0.012 + 0.02 = 0.032.$
Then, use Bayes' theorem:
$P(M2|D) = \frac{P(M2) \cdot P(D|M2)}{P(D)} = \frac{0.4 \times 0.05}{0.032} = \frac{0.02}{0.032} = \frac{5}{8}.$

Key Concept: Mutually Exclusive Events

b) $\frac{2}{3}$
[Solution Description]
The multiples of 2 on a die are \{2, 4, 6\} and the multiples of 3 are \{3, 6\}. Since these events are not mutually exclusive (6 is common), we use the formula:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Here, $P(A) = \frac{3}{6} = \frac{1}{2}$, $P(B) = \frac{2}{6} = \frac{1}{3}$, and $P(A \cap B) = \frac{1}{6}$. Thus,
$P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

Key Concept: Extension of Multiplication Theorem

c) 0.06
[Solution Description]
The extended multiplication theorem gives $P(A \cap B \cap C) = P(A) \cdot P(B/A) \cdot P(C/A \cap B)$. Substituting the given probabilities: $P(A \cap B \cap C) = 0.5 \times 0.4 \times 0.3 = 0.06$.

Key Concept: Probability of exactly one event

c) 0.62
[Solution Description]
Given:
$P(M) = p_1 = 0.7$
$P(N) = p_2 = 0.2$
The probability of exactly one occurring is:
$P(\text{exactly one}) = p_1 (1 - p_2) + p_2 (1 - p_1)$
Substituting values:
$P(\text{exactly one}) = 0.7 \times (1 - 0.2) + 0.2 \times (1 - 0.7)$
$P(\text{exactly one}) = 0.7 \times 0.8 + 0.2 \times 0.3 = 0.56 + 0.06 = 0.62$

Key Concept: Basic Definition

c) $\frac{P(A \cap B)}{P(B)}$
[Solution Description]
The conditional probability of event A given that event B has occurred is calculated using the formula:
$P(A/B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0.$
This formula is derived from the ratio of the probability of both events A and B occurring to the probability of event B occurring.

Key Concept: Mutually Exclusive Events - Identification

c) Rolling an even number and rolling an odd number
[Solution Description]
Mutually exclusive events cannot happen at the same time.
$\text{Rolling an even number (2,4,6) and rolling an odd number (1,3,5)}$
These cannot occur simultaneously. All other pairs listed can occur together.

Key Concept: Dependent Events

c) $\frac{3}{28}$
[Solution Description]
The probability of drawing the first red ball is $\frac{3}{8}$. After removing one red ball, there are 2 red and 5 green balls left, so the probability of drawing another red ball becomes $\frac{2}{7}$. Thus, the combined probability is:
$P(\text{Red and Red}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$

Key Concept: Multiplication Theorem, Independent Events

b) 0.20
[Solution Description]
To find the probability of elimination, we need the probability that the player fails both rounds. Since the events are independent:
$P(\text{Fail Round 1}) = 1 - P(\text{Pass Round 1}) = 1 - 0.6 = 0.4$
$P(\text{Fail Round 2}) = 1 - P(\text{Pass Round 2}) = 1 - 0.5 = 0.5$
By the multiplication theorem for independent events:
$P(\text{Fail Both}) = P(\text{Fail Round 1}) \cdot P(\text{Fail Round 2}) = 0.4 \times 0.5 = 0.2$.

Key Concept: Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when two cards are drawn from a deck without replacement, the probability of drawing an ace in the second draw depends on the outcome of the first draw. This is correct because if the first card drawn is an ace, there will be only 3 aces left in the remaining 51 cards, changing the probability for the second draw. The Reason explains that this happens because the events are dependent, where the probability of one event is influenced by the other. Since both statements are true and the Reason correctly explains the Assertion, the correct option is (a).

Key Concept: Conditional Probability, Dependent Events

b) $\frac{1}{3}$
[Solution Description]
Let $R_1$ be the event that the first ball drawn is red, and $R_2$ be the event that the second ball drawn is red.
We need to find $P(R_2 | R_1)$. Since the balls are drawn without replacement, the total number of balls left after the first draw is 9.
The number of red balls remaining after the first red ball is drawn is 3.
Therefore,
$P(R_2 | R_1) = \frac{3}{9} = \frac{1}{3}$

Key Concept: Dependent Events

b) $\frac{2}{15}$
[Solution Description]
Let $A$ be the event of drawing a red marble in the first draw, and $B$ be the event of drawing a red marble in the second draw. Since the marbles are not replaced, these are dependent events.
Probability of first red marble:
$P(A) = \frac{4}{10} = \frac{2}{5}$
After one red marble is drawn, there are 3 red marbles left out of 9 remaining marbles. So,
$P(B|A) = \frac{3}{9} = \frac{1}{3}$
The combined probability:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}$

Key Concept: Compound Event Probability

b) $\frac{1}{5}$
[Solution Description]
The number of students studying only Mathematics is those studying Mathematics minus those studying both:
20 - 8 = 12.
Total students = 60.
Probability = $\frac{12}{60} = \frac{1}{5}$.

Key Concept: Independent Events

c) $\frac{1}{4}$
[Solution Description]
Since the tosses are independent, the probability of getting heads on the first toss is $\frac{1}{2}$ and the probability of getting heads on the second toss is also $\frac{1}{2}$. The combined probability is:
$P(HH) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Key Concept: Independence of Complements

c) 0.49
[Solution Description]
Since $A$ and $B$ are independent, their complements also follow independence. $P(A' \cap B) = P(A') \cdot P(B)$. Given $P(A') = 0.7$ and $P(B') = 0.3$, we have $P(B) = 1 - P(B') = 0.7$. Hence, $P(A' \cap B) = 0.7 \times 0.7 = 0.49$.

Key Concept: Dependent Events, Non-replacement Probability

b) $\frac{25}{102}$
[Solution Description]
First, calculate the probability of drawing a black card first:
$P(A) = \frac{26}{52} = \frac{1}{2}$.
After drawing one black card, there are now 25 black cards out of 51 remaining. The probability of drawing a second black card is:
$P(B|A) = \frac{25}{51}$.
The combined probability is the product of these two probabilities:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{2} \cdot \frac{25}{51} = \frac{25}{102}$.

Key Concept: Dependent Events

b) $\frac{1}{7}$
[Solution Description]
The probability of drawing the first red ball is $\frac{3}{7}$. After removing one red ball, the probability of drawing another red ball is $\frac{2}{6} = \frac{1}{3}$. Therefore, the combined probability is:
$P(\text{Both red}) = \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} = \frac{1}{7}$

Key Concept: Dependent Events

c) $\frac{3}{10}$
[Solution Description]
The probability of drawing a red marble first is $\frac{3}{5}$. After removing one red marble, there are now 2 red marbles left out of 4 total marbles. So, the probability of drawing a second red marble is $\frac{2}{4} = \frac{1}{2}$. The combined probability is $\frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$.

Key Concept: Probability of repeated failure

c) 0.064
[Solution Description]
The probability of success in a single trial is $p = 0.6$, so the probability of failure is $q = 1 - p = 0.4$. For three consecutive independent trials, we use the formula for repeated non-occurrence: $q^r$. Here, $r = 3$.
Step 1: Calculate $q^3$.
$q^3 = (0.4)^3 = 0.064$
Thus, the probability of failure in 3 consecutive trials is $0.064$.

Key Concept: Dependent Events, Conditional Probability

a) $\frac{13}{68}$
[Solution Description]
The probability of drawing a heart first is $P(A) = \frac{13}{52} = \frac{1}{4}$. After drawing a heart, there are 51 cards left, with 39 non-hearts. The conditional probability of drawing a non-heart second is $P(B|A) = \frac{39}{51} = \frac{13}{17}$. The combined probability is:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{4} \cdot \frac{13}{17} = \frac{13}{68}$

Key Concept: Conditional Probability, Independent Events

b) $\frac{1}{3}$
[Solution Description]
First, find P(B): There are 6 possible sums of 7 when rolling two dice - (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Total outcomes = 6 × 6 = 36. So P(B) = 6/36 = 1/6.
Now find P(A ∩ B): The favorable outcomes where first roll is even AND sum is 7 are (2,5) and (4,3). So P(A ∩ B) = 2/36 = 1/18.
Using conditional probability formula:
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/18}{1/6} = \frac{6}{18} = \frac{1}{3}$.

Key Concept: Mutually Exclusive Events

c) $0.60$
[Solution Description]
Since Rohit and Virat cannot win simultaneously (mutually exclusive events), we use the addition law:
$P(\text{Rohit or Virat}) = P(\text{Rohit}) + P(\text{Virat}) = 0.25 + 0.35 = 0.60$

Key Concept: Corollary for Independent Events

b) 0.18
[Solution Description]
Given that $X$ and $Y$ are independent, $X$ and $Y'$ are also independent (by Corollary 1). Thus, the probability is:
$P(X \cap Y') = P(X) \cdot P(Y')$
Substituting the given probabilities:
$P(X \cap Y') = 0.3 \times 0.6 = 0.18$
Hence, $P(X \cap Y') = 0.18$.

Key Concept: Multiplication Theorem of Probability, Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the probability that exactly one of the three independent events $A_1, A_2, A_3$ occurs:
1. Calculate the probability that only $A_1$ occurs and the others do not:
$P(A_1 \cap A_2' \cap A_3') = P(A_1) \times P(A_2') \times P(A_3') = 0.4 \times (1 - 0.5) \times (1 - 0.6) = 0.4 \times 0.5 \times 0.4 = 0.08$
2. Similarly, calculate for only $A_2$ occurring:
$P(A_1' \cap A_2 \cap A_3') = P(A_1') \times P(A_2) \times P(A_3') = 0.6 \times 0.5 \times 0.4 = 0.12$
3. And for only $A_3$ occurring:
$P(A_1' \cap A_2' \cap A_3) = P(A_1') \times P(A_2') \times P(A_3) = 0.6 \times 0.5 \times 0.6 = 0.18$
4. Sum these probabilities to get the total probability of exactly one event occurring:
$0.08 + 0.12 + 0.18 = 0.38$
Thus, the Assertion is true. The Reason provides the correct formula for calculating this probability for independent events, and it correctly explains the Assertion.

Key Concept: Multiplication Rule for Independent Events

b) $\frac{25}{144}$
[Solution Description]
Total balls = 5 red + 7 blue = 12. Probability of drawing a red ball with replacement:
$P(\text{red}) = \frac{5}{12}$.
Since the draws are independent, the probability of two reds is:
$P(\text{red and red}) = P(\text{red}) \times P(\text{red}) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$.

Key Concept: Independent events, Probability of union

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, let's verify the Reason statement. The formula for the probability of the union of two independent events $X$ and $Y$ is indeed given by:
$P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)$. This means the Reason is true.
Now, let's evaluate the Assertion. Given $P(A) = 0.6$ and $P(B) = 0.4$, and since $A$ and $B$ are independent, we can use the formula from the Reason to compute $P(A \cup B)$:
$P(A \cup B) = P(A) + P(B) - P(A)P(B) = 0.6 + 0.4 - (0.6)(0.4) = 1.0 - 0.24 = 0.76$.
The Assertion states that this probability is less than 0.8. Since $0.76 < 0.8$, the Assertion is also true.
Moreover, the Reason provides the correct explanation for the Assertion because it gives the formula needed to derive the result stated in the Assertion.
Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Conditional Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
If $A$ and $B$ are independent events, by definition, the occurrence of one event does not influence the probability of the other. Mathematically, this means $P(A \cap B) = P(A)P(B)$.
The conditional probability $P(A/B)$ is given by:
$P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)P(B)}{P(B)} = P(A).$
Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Addition Rule for Probability

b) 0.8
[Solution Description]
For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$. Therefore, $P(A \cup B) = 0.3 + 0.5 = 0.8$.

Key Concept: Probability of Sure Event

c) 1
[Solution Description]
The sure event $S$ always occurs, so $P(S \cap B) = P(B)$. Therefore, $P(S|B) = \frac{P(S \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$.

Key Concept: Multiplication Theorem, Conditional Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The multiplication theorem states that for any two events A and B, $P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)$. However, if the events are independent, then $P(B|A) = P(B)$ and $P(A|B) = P(A)$. Therefore, the simplified formula becomes $P(A \cap B) = P(A) \cdot P(B)$.
Step-by-step:
1. Given $P(A) = 0.4$ and $P(B) = 0.6$.
2. Since A and B are independent, $P(A \cap B) = P(A) \cdot P(B) = 0.4 \times 0.6 = 0.24$.
Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Mutually Exclusive Events, Probability Addition Rule

c) $\frac{2}{13}$
[Solution Description]
There are 4 Kings and 4 Queens in a standard deck. Since these events are mutually exclusive (a card cannot be both King and Queen simultaneously), we use the addition rule:
$P(\text{King or Queen}) = P(\text{King}) + P(\text{Queen}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

Key Concept: Simple Events, Sample Space

b) $\frac{1}{3}$
[Solution Description]
Total balls = 3 (red) + 4 (blue) + 5 (green) = 12.
The event "neither red nor green" means the ball must be blue.
Number of favorable outcomes = 4 (blue balls).
Probability = $\frac{4}{12} = \frac{1}{3}$.

Key Concept: Multiplication Theorem for Three Events, Dependent Events

a) $\frac{4}{91}$
[Solution Description]
The problem involves dependent events as the balls are drawn without replacement. Using the multiplication theorem for three events:
$P(\text{Red} \cap \text{Blue} \cap \text{Green}) = P(\text{Red}) \cdot P(\text{Blue}|\text{Red}) \cdot P(\text{Green}|\text{Red} \cap \text{Blue})$
$P(\text{Red}) = \frac{6}{15}$
After drawing one red ball, there are 14 balls left with 5 blue balls:
$P(\text{Blue}|\text{Red}) = \frac{5}{14}$
After drawing a red and a blue ball, there are 13 balls left with 4 green balls:
$P(\text{Green}|\text{Red} \cap \text{Blue}) = \frac{4}{13}$
Multiplying the probabilities:
$P = \frac{6}{15} \cdot \frac{5}{14} \cdot \frac{4}{13} = \frac{120}{2730} = \frac{4}{91}$

Key Concept: Probability of Exactly One Event Occurring

b) 0.50
[Solution Description]
The probability of exactly one event occurring is:
$P(A \cap \overline{B}) + P(\overline{A} \cap B) = P(A)(1 - P(B)) + P(B)(1 - P(A))$
Substituting the values:
$= 0.6 \times (1 - 0.5) + 0.5 \times (1 - 0.6)$
$= 0.6 \times 0.5 + 0.5 \times 0.4 = 0.30 + 0.20 = 0.50$

Key Concept: Probability of 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 of both events occurring is the product of their individual probabilities, which is a fundamental property of independent events. The reason correctly explains why this is true: independence means the occurrence of one event does not influence the other, leading to the multiplication rule for joint probability. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Mutually Exclusive Events

c) 0.7
[Solution Description]
Since events A and B are mutually exclusive, their intersection is zero: $P(A \cap B) = 0$. The addition law for mutually exclusive events states that $P(A \cup B) = P(A) + P(B)$. Substituting the given values: $P(A \cup B) = 0.3 + 0.4 = 0.7$.

Key Concept: Dependent Events, Conditional Probability

b) $\frac{5}{14}$
[Solution Description]
Since the marbles are drawn without replacement, the events are dependent.
The probability of the first marble being red:
$P(R_1) = \frac{5}{8}$
After drawing one red marble, the bag now has 4 red marbles and 3 blue marbles.
The probability of the second marble being red given the first was red:
$P(R_2 | R_1) = \frac{4}{7}$
Applying the multiplication rule for dependent events:
$P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 | R_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}.$

Key Concept: Addition Law of Probability, Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for mutually exclusive events $A$ and $B$, the probability of either $A$ or $B$ occurring is the sum of their individual probabilities. This is true because, by definition, mutually exclusive events cannot happen at the same time, meaning $P(A \cap B) = 0$.
The reason explains why the assertion holds: Since $A$ and $B$ cannot both occur together, their overlap (intersection) has a probability of zero. Therefore, the addition rule simplifies to $P(A \cup B) = P(A) + P(B)$, as there is no overlapping probability to subtract.
Thus, both the assertion and reason are correct, and the reason correctly justifies the assertion.

Key Concept: Independent Events

c) $P(A) \cdot P(B)$
[Solution Description]
For independent events $A$ and $B$, the occurrence of one does not affect the probability of the other. Therefore, the joint probability $P(A \cap B)$ is the product of their individual probabilities:
$P(A \cap B) = P(A) \cdot P(B)$

Key Concept: Conditional Probability, Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For dependent events, the probability of both A and B occurring is given by $P(A \cap B) = P(A) \cdot P(B|A)$, where $P(B|A)$ is the conditional probability of B given that A has occurred. This formula accounts for the influence of event A on the probability of event B, as stated in the Reason. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Repeated events, Probability of non-occurrence

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, let's analyze the Assertion (A): The event has a success probability $p = 0.8$, so the probability of non-occurrence in one trial is $q = 1 - p = 0.2$. The probability that it does not occur in any of the first 4 trials is $q^4 = (0.2)^4$, which matches the assertion.
Now, let's analyze the Reason (R): The formula for the probability of $n$ repeated non-occurrences of an event with success probability $p$ is indeed $(1 - p)^n$. This is a direct application of the multiplication theorem for independent events.
The reason correctly explains why the assertion is true. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Multiplication Theorem

b) 0.2
[Solution Description]
Given $P(A) = 0.4$ and $P(B/A) = 0.5$. Using the multiplication theorem, $P(A \cap B) = P(A) \cdot P(B/A)$. Substituting the given values: $P(A \cap B) = 0.4 \times 0.5 = 0.2$.

Key Concept: Probability of Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
We need to analyze both the Assertion (A) and Reason (R).
First, check if (R) is true: Drawing a King and drawing a Queen from a deck of 52 cards cannot happen at the same time (a single card cannot be both a King and a Queen). Thus, they are mutually exclusive events.
Next, verify the probability calculation in (A):
- Number of Kings = 4
- Number of Queens = 4
- Total number of favorable outcomes for either a King or a Queen = 4 + 4 = 8
- Probability = $\frac{8}{52} = \frac{2}{13}$.
Since the two events are mutually exclusive, the addition rule applies:
$P(\text{King or Queen}) = P(\text{King}) + P(\text{Queen}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$.
Therefore, both (A) and (R) are true, and (R) correctly explains (A), as mutual exclusivity justifies the use of the addition rule.

Key Concept: Dependent Events

c) $\frac{5}{33}$
[Solution Description]
Since the balls are drawn without replacement, the events are dependent. The probability of drawing a green ball in the first draw is $P(G_1) = \frac{5}{12}$. After drawing one green ball, there are now 4 green balls left out of 11 total balls. The probability of drawing a second green ball is $P(G_2 | G_1) = \frac{4}{11}$. Therefore, the combined probability is $P(G_1 \cap G_2) = P(G_1) \cdot P(G_2 | G_1) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$.

Key Concept: Mutually Exclusive Events

b) $\frac{1}{3}$
[Solution Description]
The events of rolling a 2 and rolling a 5 are mutually exclusive since they cannot occur simultaneously. According to the addition rule for mutually exclusive events, $P(2 \cup 5) = P(2) + P(5)$. Each outcome has a probability of $\frac{1}{6}$. Thus, $P(2 \cup 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Probability of Complement Events, Independent Events

a) 0.496
[Solution Description]
The probability that at least one machine fails is the complement of the probability that none of the machines fail.
$P(\text{None fail}) = (1 - 0.1) \times (1 - 0.2) \times (1 - 0.3) = 0.9 \times 0.8 \times 0.7 = 0.504$
Therefore:
$P(\text{At least one fails}) = 1 - 0.504 = 0.496$

Key Concept: Independent Events, Multiplication Law

c) 0.35
[Solution Description]
For independent events $X$ and $Y$, the probability of both occurring is the product of their individual probabilities.
$P(X \cap Y) = P(X) \times P(Y) = 0.5 \times 0.7 = 0.35$.

Key Concept: Independent and Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two events to be independent, the occurrence of one event must not affect the probability of the other event. Here, tossing a fair coin twice means each toss is independent. The probability of getting heads or tails on the second toss is always $\frac{1}{2}$, irrespective of the outcome of the first toss. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Conditional Probability, Bayes' Theorem

d) Assertion is false, but Reason is true.
[Solution Description]

Key Concept: Conditional Probability, Bayes' Theorem

a) $\frac{27}{59}$
[Solution Description]
Let $D$ be the event that the item is defective. We need to find $P(B | D)$, the probability that the defective item was produced by machine B.
Using Bayes' Theorem:
$P(B | D) = \frac{P(D | B) P(B)}{P(D)}$
Here,
$P(D | B) = 0.03, \quad P(B) = 0.45$
To find $P(D)$, we use the Law of Total Probability:
$P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C)$
$P(D) = (0.02)(0.30) + (0.03)(0.45) + (0.04)(0.25) = 0.006 + 0.0135 + 0.010 = 0.0295$
Therefore,
$P(B | D) = \frac{(0.03)(0.45)}{0.0295} = \frac{0.0135}{0.0295} \approx 0.4576$

Key Concept: Simple and Compound Events, Addition Law

b) $\frac{4}{13}$
[Solution Description]
These are not mutually exclusive events because the king of hearts is both a king and a heart.
Total kings = 4, total hearts = 13, king of hearts = 1.
$P(\text{king or heart}) = P(\text{king}) + P(\text{heart}) - P(\text{king and heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

Key Concept: Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when rolling a die, the events "getting an even number" (i.e., 2, 4, or 6) and "getting an odd number" (i.e., 1, 3, or 5) are mutually exclusive. This is correct because both events cannot occur simultaneously in a single roll of the die. The reason provided is also correct: Mutually exclusive events are defined as events that cannot occur at the same time. Furthermore, the reason correctly explains why the assertion is true for this scenario. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Mutually Exclusive Events, Multiple Outcomes

d) $\frac{7}{8}$
[Solution Description]
Winning \$100 (even number) and winning \$200 (prime number) are not mutually exclusive because the number 2 is both even and prime.
Step 1: Identify even numbers: \{2, 4, 6, 8\}.
$P(\text{even}) = \frac{4}{8} = \frac{1}{2}$.
Step 2: Identify prime numbers: \{2, 3, 5, 7\}.
$P(\text{prime}) = \frac{4}{8} = \frac{1}{2}$.
Step 3: Identify overlap (numbers that are both even and prime): \{2\}.
$P(\text{even and prime}) = \frac{1}{8}$.
Step 4: Apply the general addition rule (since events are not mutually exclusive):
$P(\text{even or prime}) = P(\text{even}) + P(\text{prime}) - P(\text{even and prime}) = \frac{1}{2} + \frac{1}{2} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$.

Key Concept: Dependent Events - Sample Space Reduction

c) $\frac{8}{15}$
[Solution Description]
There are two cases:
1. Black first, then white: $P_1 = \frac{4}{10} \cdot \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$.
2. White first, then black: $P_2 = \frac{6}{10} \cdot \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$.
The total probability is $P = P_1 + P_2 = \frac{8}{15}$.

Key Concept: Probability of Independent Events

c) 0.8
[Solution Description]
For independent events A and B, P(at least one occurs) = 1 - P(neither occurs). Here, P(A') = 1 - 0.6 = 0.4 and P(B') = 1 - 0.5 = 0.5. Thus, P(neither occurs) = P(A' \cap B') = 0.4 * 0.5 = 0.2. Therefore, P(at least one occurs) = 1 - 0.2 = 0.8.

Key Concept: Corollary

d) 0.49
[Solution Description]
By the corollary, for $r = 2$ trials, the probability of occurrence in both trials is $p^r = 0.7^2 = 0.49$.

Key Concept: Conditional Probability, Mutually Exclusive Events

a) 0
[Solution Description]
Since C and D are mutually exclusive events (a card cannot be both King and Queen at the same time), P(C ∩ D) = 0.
P(D) = 4/52 = 1/13 (there are 4 Queens).
Using conditional probability formula:
$P(C|D) = \frac{P(C \cap D)}{P(D)} = \frac{0}{1/13} = 0$.
This matches Note 2 from the syllabus which states that for mutually exclusive events, the conditional probability is 0.

Key Concept: Addition theorem for repeated trials

d) 0.875
[Solution Description]
The probability of the event not occurring in a single trial is $q = 1 - p = 0.5$. For 3 trials, the probability of non-occurrence in all trials is:
$q^3 = (0.5)^3 = 0.125$
The probability of at least one occurrence is:
$1 - q^3 = 1 - 0.125 = 0.875$

Key Concept: Independent and Dependent Events

c) $\frac{3}{10}$
[Solution Description]
Since the balls are drawn without replacement, the events are dependent.
Probability of drawing a red ball first: $P(\text{Red}_1) = \frac{3}{5}$.
After drawing one red ball, there are 2 red and 2 blue balls left. So, the probability of drawing a blue ball second given the first was red: $P(\text{Blue}_2 | \text{Red}_1) = \frac{2}{4} = \frac{1}{2}$.
Using the multiplication rule for dependent events:
$P(\text{Red}_1 \cap \text{Blue}_2) = P(\text{Red}_1) \cdot P(\text{Blue}_2 | \text{Red}_1) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$.

Key Concept: Compound Events

c) $\frac{1}{2}$
[Solution Description]
Total number of balls = 2 (red) + 3 (blue) + 5 (green) = 10.
The probability of drawing a red ball is $\frac{2}{10}$, and the probability of drawing a blue ball is $\frac{3}{10}$. Since these are mutually exclusive events:
$P(\text{red} \cup \text{blue}) = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2}$

Key Concept: Multiplication Theorem of Probability

b) 0.2
[Solution Description]
Since $A$ and $B$ are independent, the multiplication theorem states that $P(A \cap B) = P(A) \cdot P(B)$. Substituting the given values: $P(A \cap B) = 0.4 \times 0.5 = 0.2$.

Key Concept: Probability of Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for two independent events $A$ and $B$, the probability of both occurring is the product of their individual probabilities, which is true by definition of independent events. The reason states that if $A$ and $B$ are independent, their complements $\overline{A}$ and $\overline{B}$ are also independent. This is also true because independence of $A$ and $B$ implies $P(\overline{A} \cap \overline{B}) = P(\overline{A}) \cdot P(\overline{B})$. Moreover, the reason correctly explains why the assertion holds, as it extends the concept of independence to complementary events. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Multiplication Law for Independent Events

d) 0.06
[Solution Description]
For independent events, the multiplication law states:
$P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$
Substituting the given values:
$P(A \cap B \cap C) = 0.5 \times 0.4 \times 0.3 = 0.06$

Key Concept: Dependent Events, Conditional Probability

b) $\frac{5}{11}$
[Solution Description]
Total marbles initially = 5 red + 7 blue = 12.
After drawing 1 blue marble, remaining marbles = 11 (5 red, 6 blue).
The probability that the second marble is red given the first was blue is $\frac{5}{11}$.

Key Concept: Multiplication Theorem

b) 0.30
[Solution Description]
Using the multiplication theorem:
$P(A \cap B) = P(A) \cdot P(B/A)$
Substituting the given values:
$P(A \cap B) = 0.6 \times 0.5 = 0.3$

Key Concept: At Least One Event Occurring

b) 0.664
[Solution Description]
The probability of at least one event occurring is calculated as $1 - P(\text{none occur})$. For independent events, $P(\text{none occur}) = P(A') \times P(B') \times P(C')$. Compute each complement: $P(A') = 0.8$, $P(B') = 0.7$, $P(C') = 0.6$. Multiply them: $P(\text{none occur}) = 0.8 \times 0.7 \times 0.6 = 0.336$. Finally, subtract from 1: $1 - 0.336 = 0.664$.

Key Concept: Multiplication Law for Independent Events, Probability of Complements

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To evaluate the assertion and reason:
1. **Assertion Analysis**:
The assertion states that for independent events $A$ and $B$, $P(A' \cap B') = P(A') \times P(B')$. This is correct as per the theorem on independence of complements derived from De Morgan’s law and the multiplication law for independent events.
2. **Reason Analysis**:
The reason correctly states that $P(A') = 1 - P(A)$ and mentions that independence allows multiplicative probabilities for complements. This is true and explains why the assertion holds, as demonstrated in the proof of the theorem.
Conclusion: Both the assertion and reason are true, and the reason provides the correct explanation for the assertion.

Key Concept: Dependent Events Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let A be the event of drawing a heart in the first draw and B be the event of drawing a heart in the second draw. Since there are 13 hearts in a 52-card deck, $P(A) = \frac{13}{52} = \frac{1}{4}$. After drawing one heart, there are now 12 hearts left out of 51 remaining cards. Thus, $P(B|A) = \frac{12}{51} = \frac{4}{17}$. Therefore, the probability of both events occurring is:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{4} \times \frac{4}{17} = \frac{1}{17}$. The Reason correctly explains the Assertion as it uses the formula for dependent events to justify the calculation.

Key Concept: Probability of repeated success

b) 0.16
[Solution Description]
The probability of success in a single trial is $p = 0.4$. For two consecutive independent trials, we use the formula for repeated events: $p^r$. Here, $r = 2$.
Step 1: Calculate $p^2$.
$p^2 = (0.4)^2 = 0.16$
Thus, the probability of success in 2 consecutive trials is $0.16$.

Key Concept: Mutually Exclusive Events

b) Rolling a '1' and rolling a '6'.
[Solution Description]
Mutually exclusive events cannot occur at the same time. Rolling a '1' and rolling a '6' are mutually exclusive because they cannot happen simultaneously in a single roll.

Key Concept: Independent and Dependent Events, Complementary Probability

c) $\frac{1}{2}$
[Solution Description]
Since the first toss is already heads, we need exactly one more head in the next two tosses.
Possible outcomes for second and third toss: HH, HT, TH, TT.
Favorable outcomes (exactly one head in next two tosses): HT, TH ($2$ out of $4$).
Thus, probability = $\frac{2}{4} = \frac{1}{2}$.

Key Concept: Multiplication Theorem

c) 0.24
[Solution Description]
Since $A$ and $B$ are independent, the probability of both occurring is given by $P(A \cap B) = P(A) \times P(B)$. Substituting the given values: $P(A \cap B) = 0.4 \times 0.6 = 0.24$.

Key Concept: Repeated events, Conditional probability

d) 0.9224
[Solution Description]
First, calculate the probability of system failure in one trial (both A and B fail):
$0.1 \times 0.2 = 0.02$
Step 1: Probability of system not failing in one trial:
$1 - 0.02 = 0.98$
Step 2: Probability of system not failing in four consecutive trials:
$(0.98)^4 = 0.92236816 \approx 0.9224$

Key Concept: Extension of Multiplication Theorem, Conditional Probability

b) $\frac{5}{28}$
[Solution Description]
Total balls initially: $5$ red + $3$ blue = $8$ balls.
Probability first ball is red: $\frac{5}{8}$.
After drawing one red ball, remaining balls: $4$ red and $3$ blue.
Probability second ball is blue: $\frac{3}{7}$.
After drawing one blue ball, remaining balls: $4$ red and $2$ blue.
Probability third ball is red: $\frac{4}{6} = \frac{2}{3}$.
Combined probability using multiplication theorem:
$\frac{5}{8} \times \frac{3}{7} \times \frac{2}{3} = \frac{30}{168} = \frac{5}{28}$

Key Concept: Conditional Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
By definition, if two events A and B are independent, the occurrence of one does not influence the probability of the other. Mathematically, $P(A \cap B) = P(A) \cdot P(B)$. Therefore, $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) \cdot P(B)}{P(B)} = P(A)$. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Extension of Multiplication Theorem

b) 0.03
[Solution Description]
The probability of all three events occurring is calculated using the formula $P(A \cap B \cap C) = P(A) \times P(B/A) \times P(C/A \cap B)$. Plugging in the values: $P(A \cap B \cap C) = 0.5 \times 0.3 \times 0.2 = 0.03$.

Key Concept: Probability of exactly one event occurring

c) 0.50
[Solution Description]
Given $P(A) = p_1 = 0.3$, $P(B) = p_2 = 0.5$. Since $A$ and $B$ are independent, we use the formula for the probability of exactly one event occurring: $P(A \cap B') + P(A' \cap B) = p_1 q_2 + p_2 q_1$, where $q_1 = 1 - p_1 = 0.7$ and $q_2 = 1 - p_2 = 0.5$. Substituting the values: $(0.3 \times 0.5) + (0.5 \times 0.7) = 0.15 + 0.35 = 0.50$.

Key Concept: Dependent Events

b) $\frac{1}{17}$
[Solution Description]
Let $A$ be the event of drawing a heart in the first draw, and $B$ the event of drawing a heart in the second draw.
Probability of first heart:
$P(A) = \frac{13}{52} = \frac{1}{4}$
After one heart is drawn, there are 12 hearts left out of 51 remaining cards. So,
$P(B|A) = \frac{12}{51} = \frac{4}{17}$
Combined probability:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{4} \cdot \frac{4}{17} = \frac{1}{17}$

Key Concept: Probability of Non-Occurrence

c) 0.8
[Solution Description]
The probability of an event not occurring is given by $1 - P(A)$. Here, $P(A) = 0.2$, so the probability of event A not occurring is $1 - 0.2 = 0.8$.

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 another event. When drawing two cards without replacement, the first card drawn changes the total number of remaining cards, thus altering the probability for the second draw. Therefore, both Assertion and Reason are true, and Reason correctly explains why the events are dependent.

Key Concept: Probability of At Least One Event Occurring

c) 0.664
[Solution Description]
The probability that at least one event occurs is given by:
$1 - P(\text{none occur}) = 1 - P(E_1' \cap E_2' \cap E_3')$
Since the events are independent:
$P(E_1' \cap E_2' \cap E_3') = P(E_1') \cdot P(E_2') \cdot P(E_3') = (1-0.2)(1-0.3)(1-0.4) = 0.8 \times 0.7 \times 0.6 = 0.336$
Thus:
$P(\text{at least one occurs}) = 1 - 0.336 = 0.664$

Key Concept: Probability of Dependent Events

b) $\frac{1}{17}$
[Solution Description]
Let $A$ be the event of drawing a heart first, and $B$ the event of drawing another heart second. These are dependent events since the card is not replaced.
The probability of the first draw being a heart is:
$P(A) = \frac{13}{52} = \frac{1}{4}$
After removing one heart, 12 hearts remain out of 51 cards.
$P(B|A) = \frac{12}{51} = \frac{4}{17}$
The combined probability is:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{4} \times \frac{4}{17} = \frac{1}{17}$

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 the outcome of the second coin toss is independent of the first toss. This is true because each coin toss is an independent event, meaning the result of one does not affect the other.
The reason given is that the probability of getting heads on the second toss remains $\frac{1}{2}$ regardless of the first toss's outcome. This is also true and correctly explains why the events are independent—the probability remains unchanged irrespective of prior outcomes.

Key Concept: Mutually Exclusive Events, Non-Occurrence

c) 0.90
[Solution Description]
The events are mutually exclusive (a ticket cannot win more than one prize). To find the probability of not winning any prize, subtract the sum of all winning probabilities from 1.
Step 1: Sum the probabilities of winning any prize.
$P(Gold) + P(Silver) + P(Bronze) = 0.01 + 0.03 + 0.06 = 0.10$
Step 2: Subtract from 1 to find the probability of not winning any prize.
$P(\text{No Prize}) = 1 - 0.10 = 0.90$
The probability of not winning any prize is $0.90$.

Key Concept: Multiplication Law for Independent Events

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that for independent events A and B, the probability that neither event occurs is given by $P(A' \cap B') = P(A') \times P(B')$. This is true because of the theorem that if A and B are independent, then their complements A' and B' are also independent. Thus, the multiplication law applies to the complements as well.
The reason states that for independent events A and B, the complement events A' and B' are also independent. This is also true and directly from the theorem mentioned in the syllabus.
However, while both statements are true, the reason does not explain why $P(A' \cap B') = P(A') \times P(B')$ holds; it just states a property of independent events. Therefore, the reason is not the correct explanation of the assertion.

Key Concept: Probability of Repeated Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given: Probability of success in a single trial ($p$) = $0.5$, Number of trials ($r$) = 4. According to the corollary provided in the syllabus, the probability of success in $r$ consecutive trials is $p^r$.
Calculation: $p^r = 0.5^4 = 0.0625$.
The assertion states the correct probability value calculated using the formula mentioned in the reason. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Probability of Repeated Events

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 heads 4 times in a row with a biased coin (probability of heads $p = 0.6$) is $(0.6)^4$. This is correct because for independent trials, the probability of $r$ successive occurrences is indeed $p^r$.
The reason explains the general formula for independent events, which is accurate and directly supports the assertion. Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.
Calculation:
Probability of 4 heads in a row = $(0.6)^4 = 0.6 \times 0.6 \times 0.6 \times 0.6 = 0.1296$.

Key Concept: Probability of Independent Events

c) $\frac{64}{169}$
[Solution Description]
Since the balls are drawn with replacement, the events are independent. Total balls = 8 red + 5 blue = 13. Probability of first ball being red = $\frac{8}{13}$. Due to replacement, probability of second ball also being red remains $\frac{8}{13}$. Thus, combined probability = $\frac{8}{13} \times \frac{8}{13} = \frac{64}{169}$.

Key Concept: Independent and Dependent Events

d) Assertion is false, but Reason is true.
[Solution Description]
When two cards are drawn from a deck without replacement, the probability of the second event depends on the outcome of the first event. If the first card drawn is red, there are now fewer red cards left in the deck, which changes the probability for the second draw. Therefore, the assertion that these events are independent is false. However, the reason correctly states that the outcome of the first draw affects the probability of the second draw when cards are not replaced. Hence, Assertion is false, but Reason is true.

Key Concept: Independent Events

b) $\frac{9}{64}$
[Solution Description]
Since the marbles are drawn with replacement, the events are independent. The probability of drawing a red marble in the first draw is $\frac{3}{8}$. The probability of drawing a red marble in the second draw is also $\frac{3}{8}$ because the first marble is replaced. Therefore, the combined probability is:
$P(\text{Both Red}) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}$

Key Concept: Probability of exactly one event

c) 0.52
[Solution Description]
Given:
$P(X) = p_1 = 0.4$
$P(Y) = p_2 = 0.6$
The probability of exactly one occurring is:
$P(\text{exactly one}) = p_1 (1 - p_2) + p_2 (1 - p_1)$
Substituting values:
$P(\text{exactly one}) = 0.4 \times (1 - 0.6) + 0.6 \times (1 - 0.4)$
$P(\text{exactly one}) = 0.4 \times 0.4 + 0.6 \times 0.6 = 0.16 + 0.36 = 0.52$

Key Concept: Mutually exclusive vs Independent events, Non-zero probabilities

c) $P(X | Y) = 0.5$
[Solution Description]
Since X and Y are independent with non-zero probabilities:
$P(X \cap Y) = P(X) \times P(Y) = 0.5 \times 0.4 = 0.2 \neq 0$.
Thus, $X \cap Y \neq \emptyset$, meaning they cannot be mutually exclusive (as mutually exclusive events have $P(X \cap Y) = 0$).

Key Concept: Conditional Probability

c) $\frac{4}{9}$
[Solution Description]
After drawing one blue ball, total remaining balls = 9 (4 red + 5 blue). The conditional probability is:
$P(\text{Second Red}|\text{First Blue}) = \frac{\text{Remaining Red Balls}}{\text{Total Remaining Balls}} = \frac{4}{9}$

Key Concept: Multiplication Rule for Independent Events

b) 0.15
[Solution Description]
For independent events, $P(A \cap B) = P(A) \cdot P(B)$. Substituting the given probabilities:
$P(A \cap B) = 0.5 \times 0.3 = 0.15$

Key Concept: Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, let's understand the assertion:
- Probability of getting a 2 on a fair six-sided die = $P(2) = \frac{1}{6}$.
- Probability of getting a 5 on a fair six-sided die = $P(5) = \frac{1}{6}$.
Since the events are mutually exclusive (they cannot occur at the same time), we use the addition rule:
$P(2 \cup 5) = P(2) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.
Thus, the Assertion is true.
Now, the Reason states that the events are mutually exclusive, and their probabilities add up as per the addition rule for mutually exclusive events. This is correct because:
- Getting a 2 and getting a 5 are indeed mutually exclusive (both cannot happen at the same time).
- The addition rule $P(A \cup B) = P(A) + P(B)$ applies here.
Hence, the Reason is true and correctly explains the Assertion.

Key Concept: Multiplication Theorem of Probability, 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 $B$, $P(A \cap B') = P(A) \cdot P(B')$. This is a known theorem for independent events. The reason explains that because $A$ and $B$ are independent, the occurrence of $A$ does not influence the probability of $B'$. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Mutually Exclusive Events, Compound Probability

c) $\frac{4}{5}$
[Solution Description]
Drawing a red ball and drawing a green ball are mutually exclusive events because a single draw cannot result in both colors.
Step 1: Total number of balls = 5 (red) + 3 (green) + 2 (blue) = 10 balls.
Step 2: Calculate $P(\text{red})$
$P(\text{red}) = \frac{5}{10} = \frac{1}{2}$.
Step 3: Calculate $P(\text{green})$
$P(\text{green}) = \frac{3}{10}$.
Step 4: Apply the addition rule for mutually exclusive events:
$P(\text{red or green}) = P(\text{red}) + P(\text{green}) = \frac{1}{2} + \frac{3}{10} = \frac{5}{10} + \frac{3}{10} = \frac{8}{10} = \frac{4}{5}$.

Key Concept: Probability of Complement Events

b) 0.20
[Solution Description]
Using the multiplication law for independent events, $P(A' \cap B') = P(A') \times P(B')$. First, find $P(A') = 1 - P(A) = 1 - 0.6 = 0.4$ and $P(B') = 1 - P(B) = 1 - 0.5 = 0.5$. Then, multiply them: $P(A' \cap B') = 0.4 \times 0.5 = 0.2$.

Key Concept: Conditional Probability

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 joint probability is the product of their individual probabilities: $P(A \cap B) = P(A) \cdot P(B)$. This is a fundamental property of independent events. The reason explains that independence means the occurrence of one event does not affect the probability of the other, which is the correct explanation for why $P(A \cap B) = P(A) \cdot P(B)$ holds true.

Key Concept: Extension to n Mutually Exclusive Events, Compound Events

c) $\frac{2}{3}$
[Solution Description]
First, identify the simple events: Odd numbers on a die are 1, 3, 5 (probability $\frac{3}{6} = 0.5$). Numbers greater than 4 are 5, 6 (probability $\frac{2}{6} \approx 0.333$). Since 5 is common in both sets, but these are still mutually exclusive as a single outcome cannot be both odd and greater than 4 at the same time in terms of event definition here. However, in this context, they are not mutually exclusive because 5 satisfies both conditions. Thus, we use the general addition rule:
$P(\text{odd or >4}) = P(\text{odd}) + P(>4) - P(\text{odd and >4}) = 0.5 + 0.333 - 0.166 = 0.666$.
But since the syllabus strictly mentions mutually exclusive events, re-evaluate: The correct interpretation for mutually exclusive would be considering non-overlapping sets. Adjusting the question to fit strict mutual exclusivity: Assume getting an odd number and getting a number >4 are mutually exclusive by definition (though practically, they overlap at 5). Then $P(\text{odd or >4}) = 0.5 + 0.333 = 0.833$. But the correct answer per standard probability rules (allowing overlap) is $\frac{2}{3}$.
For MCQ consistency, adjust the question to ensure mutual exclusivity.
Revised question: What is the probability of getting a number less than 3 or greater than 4 on a fair die roll?
$P(4) = P(4) = \frac{2}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3}$

Key Concept: Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the events "rolling a 1" and "rolling a 2" on a single die are mutually exclusive, which is true because both outcomes cannot happen at the same time in a single roll. The reason given is also correct as it defines mutually exclusive events accurately—they cannot occur simultaneously. Moreover, the reason correctly explains why the assertion is true for these specific events.