Class 12 Mathematics Chapter 18 Probability (Laws of Probability)

Key Concept: Conditional Probability

b) $\frac{3}{5}$
[Solution Description]
Total car owners: $10 + 15 = 25$.
$P(\text{Woman} \mid \text{Car Owner}) = \frac{\text{Number of female car owners}}{\text{Total car owners}} = \frac{15}{25} = \frac{3}{5}$

Key Concept: Independent Events - Multiplication Rule

c) $\frac{1}{4}$
[Solution Description]
The probability of getting a head on the coin toss is $\frac{1}{2}$. The die has 3 even numbers (2, 4, 6) out of 6 possible outcomes, so the probability of rolling an even number is $\frac{3}{6} = \frac{1}{2}$. Since the events are independent, the combined probability is:
$P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Key Concept: Multiplication Theorem for 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 neither event occurring is given by $(1 - P(A))(1 - P(B))$. This is correct because the non-occurrence of both events is equivalent to the intersection of their complements, and since $A$ and $B$ are independent, their complements are also independent. Thus, $P(A' \cap B') = P(A') \cdot P(B') = (1 - P(A))(1 - P(B))$.
The Reason states that for independent events, $P(A \cap B) = P(A) \cdot P(B)$. This is a fundamental property of independent events and is true.
Additionally, the Reason correctly explains why the Assertion is true. Since the complements of independent events are also independent, the multiplication theorem applies directly to their probabilities of non-occurrence. Therefore, the Reason is the correct explanation for the Assertion.

Key Concept: Addition Theorem for Independent Events

c) 0.58
[Solution Description]
Using the addition theorem for independent events:
$P(A \cup B) = 1 - P(\overline{A}) \times P(\overline{B})$
Given P(A) = 0.4, P(B) = 0.3, so:
$P(\overline{A}) = 1 - 0.4 = 0.6$
$P(\overline{B}) = 1 - 0.3 = 0.7$
$P(A \cup B) = 1 - (0.6 \times 0.7) = 1 - 0.42 = 0.58$

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: 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: Multiplication Rule for Independent Events

a) 0.18
[Solution Description]
For independent events, $P(A \cap B) = P(A) \times P(B)$. Substituting the given values: $P(A \cap B) = 0.6 \times 0.3 = 0.18$.

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 a die is rolled, the events "getting an even number" and "getting an odd number" are mutually exclusive. This is true because these two events cannot occur at the same time (e.g., you cannot roll both 2 (even) and 3 (odd) in a single roll).
The reason states that an even number and an odd number cannot occur simultaneously in a single roll of a die. This is also true as a die can show only one face at a time, which will be either even or odd but not both.
Moreover, the reason correctly explains why the assertion is true, as it explicitly highlights the impossibility of simultaneous occurrence, which defines mutual exclusivity.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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: Addition Law of Probability: Extension to Multiple Exclusive Events

b) 0.4
[Solution Description]
Since A and C are mutually exclusive events, we can use the addition law for multiple exclusive events. The probability that either A or C wins is the sum of their individual probabilities. Therefore:
$P(A \cup C) = P(A) + P(C) = 0.1 + 0.3 = 0.4$

Key Concept: Mutually Exclusive Events, 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 for mutually exclusive events A and B, the probability of their union is the sum of their individual probabilities, which is true as per the addition law for mutually exclusive events. The reason given is that $P(A \cap B) = 0$ for mutually exclusive events, which is also correct because such events cannot occur at the same time. Moreover, the reason correctly explains why the addition law holds for mutually exclusive events—the absence of overlap ($A \cap B$) ensures that their probabilities simply add up. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Multiplication Theorem for Three Events, Independent and Dependent 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, the joint probability $P(A \cap B \cap C)$ is the product of their individual probabilities. This is true because independence implies that the occurrence of one event does not influence the others. The reason explains the definition of independent events, which directly justifies the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: 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.

Key Concept: Conditional Probability, Multiplication Theorem

c) $\frac{5}{11}$
[Solution Description]
Total balls initially = 5 red + 7 blue = 12. After drawing one blue ball, remaining balls = 11 (5 red and 6 blue). The probability that the second ball is red given first was blue is the number of remaining red balls divided by total remaining balls: $P(R_2|B_1) = \frac{5}{11}$.

Key Concept: Dependent Events, Conditional Probability

c) 0.67
[Solution Description]
Let $A$ be the event that the first ball is red, and $B$ be the event that the second ball is blue. We need to find $P(A | B)$.
Using Bayes' theorem:
$P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}$
Calculate each part:
$P(A) = \frac{6}{10} = 0.6$
$P(B | A) = \frac{4}{9}$ (since one red ball is already drawn)
$P(B) = P(B | A)P(A) + P(B | A')P(A') = \left(\frac{4}{9} \times \frac{6}{10}\right) + \left(\frac{3}{9} \times \frac{4}{10}\right) = \frac{24}{90} + \frac{12}{90} = \frac{36}{90} = 0.4$
Now substitute back:
$P(A | B) = \frac{\frac{4}{9} \times \frac{6}{10}}{0.4} = \frac{\frac{24}{90}}{0.4} = \frac{24}{36} = \frac{2}{3} \approx 0.6667$.

Key Concept: Conditional Probability, Dependent Events

c) $\frac{2}{13}$
[Solution Description]
First, calculate the probability of drawing a white ball first:
$P(A) = \frac{8}{14} = \frac{4}{7}$.
After drawing one white ball, there are 7 white and 6 black balls left. The probability of drawing a second white ball is:
$P(B|A) = \frac{7}{13}$.
After drawing two white balls, there are 6 white and 6 black balls left. The probability of drawing a third white ball is:
$P(C|A \cap B) = \frac{6}{12} = \frac{1}{2}$.
The combined probability is the product of these three probabilities:
$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B) = \frac{4}{7} \cdot \frac{7}{13} \cdot \frac{1}{2} = \frac{28}{182} = \frac{2}{13}$.

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: Probability of Failure

a) 0.210
[Solution Description]
The probability of all three events failing is given by $(1 - p_1)(1 - p_2)(1 - p_3) = (1 - 0.3)(1 - 0.4)(1 - 0.5) = 0.7 \times 0.6 \times 0.5 = 0.21$.

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: Dependent Events

a) $\frac{11}{221}$
[Solution Description]
There are 12 face cards in a standard deck. The probability of first card being a face card is $P(A) = \frac{12}{52} = \frac{3}{13}$. If the first card is a face card, then 11 face cards remain out of 51 cards. So, $P(B|A) = \frac{11}{51}$. Therefore, $P(A \cap B) = P(A) \cdot P(B|A) = \frac{3}{13} \times \frac{11}{51} = \frac{33}{663} = \frac{11}{221}$.

Key Concept: Addition Law of Probability, Mutually Exclusive Events

c) 0.475
[Solution Description]
First, list all possible pairs and their sums:
(1,2)=3, (1,4)=5, (1,6)=7, (1,8)=9,
(3,2)=5, (3,4)=7, (3,6)=9, (3,8)=11,
(5,2)=7, (5,4)=9, (5,6)=11, (5,8)=13.
Favorable outcomes (sum >7): (1,8), (3,6), (3,8), (5,4), (5,6), (5,8).
Calculate probabilities:
$P(1,8)=0.4 \times 0.25=0.1$,
$P(3,6)=0.3 \times 0.25=0.075$,
$P(3,8)=0.3 \times 0.25=0.075$,
$P(5,4)=0.3 \times 0.25=0.075$,
$P(5,6)=0.3 \times 0.25=0.075$,
$P(5,8)=0.3 \times 0.25=0.075$.
Total probability = $0.1+0.075+0.075+0.075+0.075+0.075=0.475$.

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, 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: 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: Dependent Events, Conditional Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for dependent events $A$ and $B$, the joint probability is $P(A \cap B) = P(B) \cdot P(A|B)$. This is correct because it aligns with the definition of conditional probability for dependent events. The reason claims that dependent events cannot be independent events, which is also true, as independence implies $P(A|B) = P(A)$, making them no longer dependent. The reason correctly explains why the assertion holds in the context of dependent events.

Key Concept: Multiplication Law for Independent Events

c) $\frac{1}{16}$
[Solution Description]
Each toss is an independent event with $P(\text{Heads}) = 0.5$. Using the multiplication law:
$P(\text{All Heads}) = P(\text{H}) \times P(\text{H}) \times P(\text{H}) \times P(\text{H}) = 0.5^4$
Calculating:
$0.5^4 = 0.0625$

Key Concept: Independent Events Probability

b) 0.20
[Solution Description]
For independent events, the probability of both occurring is the product of their individual probabilities. So, $P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2$.

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: Multiplication Theorem of 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 probability of their intersection is the product of their individual probabilities. This is correct as per the multiplication theorem for independent events. The reason provided correctly defines the condition for independent events and accurately explains the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: Conditional Probability, Multiple Events

b) $\frac{4}{51}$
[Solution Description]
Given:
- Total cards = 52, number of aces = 4.
- First card is not an ace, so remaining cards = 51, remaining aces = 4.
- We need $P(\text{Second is ace} | \text{First is not ace})$.
The probability is:
$P(\text{Second is ace} | \text{First is not ace}) = \frac{\text{Number of aces left}}{\text{Total cards left}} = \frac{4}{51}.$

Key Concept: Multiplication Theorem of Probability for n Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the multiplication theorem for independent events, which is correct as per the given syllabus. The reason correctly explains why the theorem holds because independence implies no mutual influence on probabilities between events. Thus, both assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Independence of Complements

c) $0.5$
[Solution Description] To find the probability of exactly one event occurring, we calculate:
$P(\text{Exactly one}) = P(A \cap B') + P(A' \cap B)$
Since $A$ and $B$ are independent, their complements are also independent. Thus,
$P(A \cap B') = P(A) \cdot P(B') = 0.3 \times (1 - 0.5) = 0.15$
$P(A' \cap B) = P(A') \cdot P(B) = (1 - 0.3) \times 0.5 = 0.35$
Therefore,
$P(\text{Exactly one}) = 0.15 + 0.35 = 0.5$

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: Addition Law of Probability, Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct formula for the probability of the union of three mutually exclusive events, which is indeed $P(A) + P(B) + P(C)$. The reason correctly explains this by stating that the addition law applies here as the intersection of any two events is zero ($P(A \cap B) = P(B \cap C) = P(A \cap C) = 0$), making the events mutually exclusive. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Probability of exactly one of two independent events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, calculate $P(A') = 1 - P(A) = 1 - 0.6 = 0.4$ and $P(B') = 1 - P(B) = 1 - 0.4 = 0.6$. Using the formula for the probability of exactly one of the two events occurring, we substitute the values: $P(A \cap B') + P(A' \cap B) = P(A)P(B') + P(B)P(A') = (0.6)(0.6) + (0.4)(0.4) = 0.36 + 0.16 = 0.52$. Thus, the Assertion is true. The Reason provides the correct formula for calculating the probability of exactly one of two independent events occurring, which is also true and correctly explains the Assertion.

Key Concept: Assertion-Reason

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 either a 1 or a 2 when rolling a die is $\frac{1}{3}$. This is correct because:
- Probability of getting a 1 ($P(1)$) = $\frac{1}{6}$
- Probability of getting a 2 ($P(2)$) = $\frac{1}{6}$
Since these are mutually exclusive events (as per the Reason), the addition rule for exclusive events applies:
$P(1 \cup 2) = P(1) + P(2) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$.
The Reason correctly identifies that the two events are mutually exclusive, which justifies the use of the addition rule.

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: 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 non-occurrence of repeated events

b) $(0.3)^3 = 0.027$
[Solution Description]
Probability of failure in a single trial, $q = 1 - p = 1 - 0.7 = 0.3$. For $r = 3$ consecutive failures, the probability is $q^r = (0.3)^3 = 0.027$.

Key Concept: Conditional Probability, Independent Events

b) $\frac{18}{67}$
[Solution Description]
Let $D$ be the event that the person has the disease, and $T^+$ be the event that the test is positive.
We need to find $P(D | T^+)$. Using Bayes' Theorem:
$P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)}$
Here,
$P(T^+ | D) = 1 - P(\text{false negative}) = 1 - 0.10 = 0.90$
$P(D) = 0.02$
Now, $P(T^+)$ can be found using the Law of Total Probability:
$P(T^+) = P(T^+ | D) P(D) + P(T^+ | D') P(D')$
$P(T^+ | D') = P(\text{false positive}) = 0.05$
$P(D') = 1 - P(D) = 0.98$
So,
$P(T^+) = (0.90)(0.02) + (0.05)(0.98) = 0.018 + 0.049 = 0.067$
Therefore,
$P(D | T^+) = \frac{(0.90)(0.02)}{0.067} = \frac{0.018}{0.067} \approx 0.2687$

Key Concept: Simple Events, Mutually Exclusive Events

b) $\frac{1}{4}$
[Solution Description]
First, identify simple events: Prime numbers on an eight-sided die are 2, 3, 5, 7 (4 outcomes). Composite numbers are 4, 6, 8 (3 outcomes). Even numbers are 2, 4, 6, 8 (4 outcomes). However, "composite or even" includes all even numbers plus odd composites - but there are no odd composites on an eight-sided die. So it's just even numbers.
Probability of first roll being prime: $P(A) = \frac{4}{8} = \frac{1}{2}$
Probability of second roll being even: $P(B) = \frac{4}{8} = \frac{1}{2}$
Since these are independent events, combined probability is:
$P(A \text{ and } B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Key Concept: Probability of Exactly One of Two Events

b) 0.5
[Solution Description]
The probability that exactly one of the two events occurs is given by $P(A \cap B') + P(A' \cap B)$. Since A and B are independent, $P(A \cap B') = P(A)(1 - P(B)) = 0.6 \times 0.5 = 0.3$, and $P(A' \cap B) = (1 - P(A))P(B) = 0.4 \times 0.5 = 0.2$. Adding these gives $0.3 + 0.2 = 0.5$.

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: 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 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: Difference Between Mutually Exclusive and Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two independent events $A$ and $B$ with $P(A) > 0$ and $P(B) > 0$, the probability of their intersection is $P(A \cap B) = P(A) \times P(B)$. This means $P(A \cap B) \neq 0$. However, for mutually exclusive events, $P(A \cap B) = 0$ because they cannot occur simultaneously. Therefore, if events are independent and have non-zero probabilities, they cannot be mutually exclusive. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Independent Events

b) $P(A) \cdot P(B)$
[Solution Description]
Since A and B are independent, the probability of both events occurring is given by $P(A \cap B) = P(A) \cdot P(B)$.

Key Concept: Combined probability of repeated events and non-occurrence

d) $(0.6)^2 \times (0.4)^3 = 0.02304$
[Solution Description]
Probability of heads, $p = 0.6$, and tails, $q = 1 - 0.6 = 0.4$. The probability of 2 heads followed by 3 tails is $p^2 \times q^3 = (0.6)^2 \times (0.4)^3 = 0.36 \times 0.064 = 0.02304$.

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: Dependent Events, Conditional Probability

c) $\frac{7}{11}$
[Solution Description]
Since the first ball drawn is red and not replaced, the bag now has 4 red balls and 7 green balls, making a total of 11 balls left. The probability of drawing a green ball in the second draw is therefore $\frac{7}{11}$ because there are 7 green balls out of the remaining 11 balls.

Key Concept: Conditional Probability, Property Application

b) 0.4
[Solution Description]
First find P(A ∩ B) using the general addition rule:
P(A∪B) = P(A) + P(B) - P(A ∩ B)
0.7 = 0.4 + 0.5 - P(A ∩ B)
P(A ∩ B) = 0.9 - 0.7 = 0.2
Now apply the conditional probability formula:
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4$
Notice this equals P(A), illustrating Property 2 where if events are independent, P(A|B) = P(A).

Key Concept: Independent events, Probability of at least one event occurring

a) 0.664
[Solution Description]
To find the probability that at least one component fails, we first calculate the probability that all components function correctly using the independence property:
$P(\text{all work}) = P(A) \times P(B) \times P(C) = 0.8 \times 0.7 \times 0.6 = 0.336$.
Then, the probability of at least one failure is:
$1 - P(\text{all work}) = 1 - 0.336 = 0.664$.

Key Concept: Multiplication Theorem, Independent Events

b) $\frac{15}{19}$
[Solution Description]
Let D be the event of defective item. Using Bayes' theorem:
$P(A|D) = \frac{P(D|A)P(A)}{P(D|A)P(A) + P(D|B)P(B)} = \frac{0.05 \times 0.6}{(0.05 \times 0.6) + (0.02 \times 0.4)} = \frac{0.03}{0.03 + 0.008} = \frac{0.03}{0.038} = \frac{15}{19}$.

Key Concept: Simple Event

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that rolling a die and getting a '5' is a simple event. This is true because a simple event is defined as a single outcome of an experiment, such as getting a specific number on a die. The reason states that a simple event represents only one possible outcome of an experiment, which is the correct definition of a simple event and correctly explains why the assertion is true.

Key Concept: Probability of exactly one event occurring

b) 0.16
[Solution Description]
The probability that only event $A$ occurs is given by $P(A \cap B')$. Since $A$ and $B$ are independent, $P(A \cap B') = P(A) \cdot P(B') = P(A) \cdot (1 - P(B))$. Substituting the given probabilities: $0.4 \times (1 - 0.6) = 0.4 \times 0.4 = 0.16$.

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 Non-Occurrence

b) 0.612
[Solution Description]
The probability that none fail is the product of their individual success probabilities. Given failure probabilities $p_1 = 0.1$, $p_2 = 0.15$, $p_3 = 0.2$, the success probabilities are $1 - p_1 = 0.9$, $1 - p_2 = 0.85$, and $1 - p_3 = 0.8$. Thus:
$P(\text{None fail}) = (1 - p_1)(1 - p_2)(1 - p_3) = 0.9 \times 0.85 \times 0.8 = 0.612$

Key Concept: Independent Events, Multiplication Theorem for Three Events

b) $\frac{3}{8}$
[Solution Description]
Since the coin tosses are independent events, we can use the multiplication theorem for independent events.
There are three possible sequences for getting exactly two heads and one tail: HHT, HTH, THH.
Each sequence has a probability of:
$P(H \cap H \cap T) = P(H) \cdot P(H) \cdot P(T) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$
Since there are 3 such sequences:
$\text{Total Probability} = 3 \cdot \frac{1}{8} = \frac{3}{8}$

Key Concept: Addition Law for Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]

Key Concept: Mutually Exclusive Events

c) $\frac{1}{3}$
[Solution Description]
Here, getting a '3' and a '5' in a single throw are mutually exclusive events. The probability of each outcome is $\frac{1}{6}$. Using the addition rule for exclusive events:
$P(3 \cup 5) = P(3) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Independent Events, Probability of Complementary Events

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
From the theorem provided in the syllabus, if $A$ and $B$ are independent events, then $A'$ and $B'$ are also independent (as per part (c) of the theorem). Hence, Assertion (A) is true. The Reason (R) states the correct formula for the probability of the union of two independent events, which is derived from the inclusion-exclusion principle and uses the fact that $P(A \cap B) = P(A)P(B)$ for independent events. However, Reason (R) does not explain why $A'$ and $B'$ are independent, it just provides a property of independent events. Therefore, both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

Key Concept: Independent Events, Addition Law

c) 0.7
[Solution Description]
Since A and B are independent, $P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2$.
Using addition law for non-mutually exclusive events:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$.

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 P(A \cap B) equals the product of their individual probabilities P(A) and P(B), which is a fundamental property of independent events. The reason correctly explains the assertion by stating that independence implies the occurrence of one event does not influence the probability of the other. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Independent Events

c) 0.54
[Solution Description]
The probability of exactly one of two independent events occurring is given by $P(A)(1 - P(B)) + P(B)(1 - P(A))$. Here, $P(A) = 0.6$ and $P(B) = 0.3$. So, the probability is $0.6 \times (1 - 0.3) + 0.3 \times (1 - 0.6) = 0.6 \times 0.7 + 0.3 \times 0.4 = 0.42 + 0.12 = 0.54$.

Key Concept: Independent Events, Conditional Probability

b) No, because $P(F \cap B) \neq P(F) \cdot P(B)$
[Solution Description]
Let $F$ be the event a student plays football ($P(F) = 0.4$) and $B$ be the event a student plays basketball ($P(B) = 0.3$). Given $P(F \cap B) = 0.15$. For independence, check if $P(F \cap B) = P(F) \cdot P(B)$. Here, $0.4 \times 0.3 = 0.12 \neq 0.15$, so the events are dependent.

Key Concept: Addition Law for Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the addition law for mutually exclusive events: $P(A \cup B) = P(A) + P(B)$, which is correct. The reason explains that mutually exclusive events cannot occur simultaneously, which is also correct. Further, the reason directly justifies the assertion because if two events cannot occur together, their combined probability is simply the sum of their individual probabilities. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Addition Law of Probability, Mutually Exclusive Events

c) 0.4
[Solution Description]
Using the principle of inclusion-exclusion for non-mutually exclusive events:
$P(\text{Football} \cup \text{Basketball}) = P(\text{Football}) + P(\text{Basketball}) - P(\text{Both})$.
Substituting values: $0.4 + 0.3 - 0.1 = 0.6$.
Probability of playing neither = $1 - 0.6 = 0.4$.

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 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: Dependent Events Probability

a) $\frac{11}{221}$
[Solution Description]
There are 12 face cards in a deck (4 Jacks, 4 Queens, 4 Kings). The probability of drawing the first face card is $P(A) = \frac{12}{52} = \frac{3}{13}$. After drawing one face card, there are now 11 face cards left out of 51 remaining cards. So, the probability of drawing a second face card is $P(B|A) = \frac{11}{51}$. The combined probability is:
$P(A \cap B) = P(A) \cdot P(B|A) = \frac{3}{13} \cdot \frac{11}{51} = \frac{33}{663} = \frac{11}{221}.$

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: 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: Independence of Complements

b) 0.36
[Solution Description]
Since $A$ and $B$ are independent, $P(A \cap B') = P(A) \cdot P(B')$. Here, $P(B') = 1 - P(B) = 1 - 0.4 = 0.6$. Thus, $P(A \cap B') = 0.6 \times 0.6 = 0.36$.

Key Concept: Independent vs Dependent

c) Drawing two cards from a deck without replacement
[Solution Description]
Dependent events occur when the outcome of one event affects the probability of the other. Drawing two cards from a deck without replacement is dependent because the first card drawn changes the composition of the deck for the second draw.

Key Concept: Independent and Dependent Events, Multiplication Rule for 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, the occurrence of one does not affect the probability of the other. This is the definition of independent events. The reason provides the mathematical condition for independence: $P(A \cap B) = P(A) \cdot P(B)$, which directly supports the assertion as it quantifies the lack of influence between the events. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Multiplication Theorem of Probability, 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 if two events A and B are independent, then $P(B|A) = P(B)$. This is true because, by definition, for independent events, the occurrence of one event does not influence the probability of the other event. The reason provided is also true and correctly explains the assertion, as it gives the definition of independent events which directly leads to the assertion being valid.

Key Concept: Mutually Exclusive Events

d) $\frac{1}{3}$
[Solution Description]
Since rolling a 2 and rolling a 5 are mutually exclusive events, we use the addition rule for exclusive events:
$P(2 \cup 5) = P(2) + P(5)$. For a fair die, each number has an equal probability of $\frac{1}{6}$. Thus:
$P(2 \cup 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}.$

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 two independent events $A$ and $B$, the probability of their intersection is the product of their individual probabilities. This is true because, by definition, independent events satisfy this condition. The reason correctly explains why the assertion holds: independence means the occurrence of one event has no impact on the probability of the other, which mathematically translates to $P(A \cap B) = P(A) \cdot P(B)$. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Independent Events

a) $\frac{1}{169}$
[Solution Description]
Since the draws are independent, the probability of drawing a king in each draw remains the same.
$P(\text{First king}) = \frac{4}{52} = \frac{1}{13}$
$P(\text{Second king}) = \frac{4}{52} = \frac{1}{13}$
The combined probability:
$P(\text{Both kings}) = \frac{1}{13} \cdot \frac{1}{13} = \frac{1}{169}$

Key Concept: Independent vs Dependent Events

a) $\frac{25}{144}$
[Solution Description]
Since the marbles are drawn with replacement, the events are independent. The probability of drawing a green marble each time is $\frac{5}{12}$. The combined probability is $\frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$.

Key Concept: Simple and Compound Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that rolling a 4 on a fair six-sided die is a simple event. This is true because a simple event refers to an event with a single outcome, which in this case is rolling a 4. The reason provided explains that a simple event describes a single outcome, which aligns perfectly with the definition of a simple event. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

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: Mutually Exclusive Events, Complex Probability

c) 0.75
[Solution Description]
Since the events are mutually exclusive, the probability of winning at least one prize is the sum of individual probabilities:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) = 0.25 + 0.35 + 0.15 = 0.75$

Key Concept: Repeated events, Probability of success and failure

b) 0.488
[Solution Description]
To find the probability of at least one defective item in three trials, we calculate the complement of no defective items in all three trials. The probability of non-defective item in one trial is $p = 0.8$.
Step 1: Calculate probability of all three items being non-defective:
$p^3 = (0.8)^3 = 0.512$
Step 2: Subtract from 1 to get probability of at least one defective item:
$1 - p^3 = 1 - 0.512 = 0.488$

Key Concept: Extension to Multiple Exclusive Events

d) 0.80
[Solution Description]
Since only one player can win at a time, the events are mutually exclusive.
$P(X \cup Y \cup Z) = P(X) + P(Y) + P(Z) = 0.25 + 0.35 + 0.20 = 0.80$

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: Probability of exactly one event

c) 0.50
[Solution Description]
Given:
$P(A) = p_1 = 0.3$
$P(B) = p_2 = 0.5$
Since $A$ and $B$ are independent, the probability of exactly one occurring is:
$P(\text{exactly one}) = p_1 (1 - p_2) + p_2 (1 - p_1)$
Substituting values:
$P(\text{exactly one}) = 0.3 \times (1 - 0.5) + 0.5 \times (1 - 0.3)$
$P(\text{exactly one}) = 0.3 \times 0.5 + 0.5 \times 0.7 = 0.15 + 0.35 = 0.50$

Key Concept: Dependent Events - Conditional Probability

b) $\frac{1}{17}$
[Solution Description]
The probability of drawing the first heart is $\frac{13}{52} = \frac{1}{4}$. Since we do not replace the card, the probability of drawing a second heart becomes $\frac{12}{51} = \frac{4}{17}$. The combined probability is $P = \frac{1}{4} \cdot \frac{4}{17} = \frac{1}{17}$.

Key Concept: Addition Law of Probability, Mutually Exclusive Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For mutually exclusive events, the addition law simplifies to the sum of individual probabilities:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) = 0.4 + 0.3 + 0.2 = 0.9$
The Assertion calculates the correct probability using the addition law for mutually exclusive events, and the Reason provides the correct mathematical rule. Additionally, the Reason correctly explains why the Assertion holds true.

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

b) $\frac{1}{3}$
[Solution Description]
These are mutually exclusive events since both cannot occur at the same time. Using the addition rule for exclusive events:
$P(3 \cup 5) = P(3) + P(5)$.
Since each outcome has a probability of $\frac{1}{6}$, we get:
$P(3 \cup 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

Key Concept: Independent Events

c) 0.06
[Solution Description]
Since $A$, $B$, and $C$ are independent, their joint probability is given by:
$P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C) = 0.5 \cdot 0.4 \cdot 0.3 = 0.06$.

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: Probability of Non-Occurrence

b) 0.09
[Solution Description]
The probability of non-occurrence of A in one trial is $1 - P(A) = 0.3$. For two independent trials, the combined probability of non-occurrence is:
$(1 - P(A))^2 = 0.3 \times 0.3 = 0.09$

Key Concept: Extension of Multiplication Theorem

a) 0.941094
[Solution Description]
The probability that each machine produces a non-defective bolt is $(1 - 0.01)$, $(1 - 0.02)$, and $(1 - 0.03)$ respectively. Since these events are independent, the combined probability is $(1 - 0.01) \times (1 - 0.02) \times (1 - 0.03) = 0.99 \times 0.98 \times 0.97 = 0.941094$.

Key Concept: Independent Events

b) $\frac{1}{4}$
[Solution Description]
Since the two tosses are independent events, the probability of getting heads on both tosses is the product of the probabilities of getting heads on each toss. The probability of heads on a single toss is $\frac{1}{2}$. Therefore, $P(\text{Heads on both tosses}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Key Concept: Dependent Events - Basic Probability Calculation

b) $\frac{2}{15}$
[Solution Description]
The probability of drawing a red marble in the first draw is $\frac{4}{10} = \frac{2}{5}$. After one red marble is drawn, there are 3 red marbles left out of 9 total marbles. The probability of drawing another red marble is $\frac{3}{9} = \frac{1}{3}$. Therefore, the combined probability is:
$P(\text{both red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}.$

Key Concept: Compound Event, Independent and Dependent Events

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that for independent events $A$ and $B$, $P(A \cap B) = P(A) + P(B)$. This is incorrect because for independent events, the correct formula is $P(A \cap B) = P(A) \cdot P(B)$, as given in the reason. Therefore, the assertion is false, but the reason is true.

Key Concept: Dependent Events, Conditional Probability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the probability of drawing two red balls in succession without replacement:
1. Probability of first red ball: $P(A) = \frac{5}{8}$ (since there are 5 red balls out of 8 total).
2. After drawing one red ball, there are now 4 red balls left out of 7 remaining balls. Thus, the conditional probability of drawing a second red ball is $P(B|A) = \frac{4}{7}$.
3. The combined probability is $P(A \cap B) = P(A) \cdot P(B|A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.
The Reason correctly explains why the probability changes in dependent events when sampling without replacement.

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: 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: Dependent Events

c) $\frac{4}{15}$
[Solution Description]
Probability of first marble being white, $P(A) = \frac{4}{10} = \frac{2}{5}$. After removing one white marble, there are 6 black marbles left out of 9. So, $P(B|A) = \frac{6}{9} = \frac{2}{3}$. Therefore, $P(A \cap B) = P(A) \cdot P(B|A) = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15}$.

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

b) $\frac{1}{3}$
[Solution Description]
Since rolling a 1 and rolling a 6 are mutually exclusive events (both cannot occur at the same time), we use the addition law for exclusive events:
$P(1 \cup 6) = P(1) + P(6)$
The probability of rolling a 1 is $\frac{1}{6}$ and the probability of rolling a 6 is also $\frac{1}{6}$. Therefore:
$P(1 \cup 6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

Key Concept: Addition Law for Exclusive Events

d) 0.9
[Solution Description]
Since $A$, $B$, and $C$ are mutually exclusive, we use the addition law:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) = 0.3 + 0.4 + 0.2 = 0.9$

Key Concept: Compound Events, Addition Law

b) 0.20
[Solution Description]
This is a compound event requiring the general addition rule since the events are not mutually exclusive.
Given:
$P(X) = 0.15$,
$P(Y) = 0.08$,
$P(X \cap Y) = 0.03$
The probability of winning at least one prize is:
$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.15 + 0.08 - 0.03 = 0.20$

Key Concept: Independent Events, Multiplication Rule

c) $\frac{1}{2}$
[Solution Description]
First, note that the outcome of the first toss is already given as a head. We need to find the probability of getting exactly one more head and one tail in the next two tosses. Since the coin tosses are independent, the probability of any outcome in the remaining two tosses is not affected by the first toss. The possible favorable outcomes for the next two tosses are HT or TH. The probability of each of these outcomes is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Since there are two such favorable outcomes, the total probability is $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

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

b) $\frac{1}{3}$
[Solution Description]
Probability of drawing first red ball, $P(A) = \frac{6}{10}$. After removing one red ball, remaining red balls are 5 out of total 9 balls. So, probability of second red ball given first was red, $P(B|A) = \frac{5}{9}$. Thus, $P(A \cap B) = P(A) \cdot P(B|A) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.

Key Concept: Conditional Probability Formula

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

Key Concept: Independent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] For two independent events $A$ and $B$, the probability of their intersection is the product of their individual probabilities, as given in the Assertion. The Reason correctly explains this by stating that the occurrence of one event does not influence the other. Since both statements are true and the Reason justifies the Assertion, the correct answer is a).

Key Concept: Probability of at least one event, Independent events

a) 0.976
[Solution Description]
To find the probability that the student passes at least one test, we can use the formula for the probability of at least one event occurring among independent events: $1 - (1 - p_1)(1 - p_2)(1 - p_3)$. Here, $p_1 = 0.8$, $p_2 = 0.7$, and $p_3 = 0.6$.
First, calculate $(1 - p_1) = 1 - 0.8 = 0.2$.
Next, $(1 - p_2) = 1 - 0.7 = 0.3$.
Then, $(1 - p_3) = 1 - 0.6 = 0.4$.
Now multiply these probabilities: $0.2 \times 0.3 \times 0.4 = 0.024$.
Finally, subtract this from 1: $1 - 0.024 = 0.976$.
Therefore, the probability that the student passes at least one test is 0.976.

Key Concept: Independent and Dependent Events

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For independent events, the probability of both events occurring together is the product of their individual probabilities. This is because the occurrence of one event does not influence the other. Here, the assertion correctly states the mathematical condition for independence: $P(A \cap B) = P(A) \cdot P(B)$. The reason explains why this condition holds—because the occurrence of $A$ does not affect $B$. Thus, the reason correctly explains the assertion.

Key Concept: Independent Events

b) $\frac{4}{25}$
[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 $P(R_1) = \frac{4}{10} = \frac{2}{5}$. Similarly, the probability of drawing a red marble in the second draw is $P(R_2) = \frac{4}{10} = \frac{2}{5}$. Using the multiplication rule for independent events, the probability that both marbles are red is $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$.

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 Law of Probability

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 their union is the sum of their individual probabilities, which is correct as per the addition rule for exclusive events. The reason correctly explains that since mutually exclusive events cannot occur together, their intersection has a probability of 0. Thus, both the assertion and reason are true, and the reason justifies the assertion.

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: 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: Dependent Events

a) $\frac{2}{15}$
[Solution Description]
The probability of drawing a red marble in the first draw is $P(A) = \frac{4}{10} = \frac{2}{5}$. After drawing one red marble, there are 3 red marbles left out of 9. So, the probability of drawing another red marble is $P(B|A) = \frac{3}{9} = \frac{1}{3}$. The combined probability is $P(A \cap B) = P(A) \cdot P(B|A) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}$.

Key Concept: Conditional Probability, Multiplication Theorem, 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 occurring is $P(A) \cdot P(B)$. This is true because, by definition, independent events satisfy $P(A \cap B) = P(A) \cdot P(B)$. The reason states that for independent events, $P(A/B) = P(A)$ and $P(B/A) = P(B)$. This is also true as it follows from the definition of conditional probability and independence, since $P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)P(B)}{P(B)} = P(A)$, and similarly for $P(B/A)$. Moreover, the reason correctly explains why the assertion holds, because the independence condition directly leads to the formula mentioned in the assertion. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Conditional Probability

b) 0.4
[Solution Description]
To find $P(A|B)$, we use the formula for conditional probability:
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
Substituting the given values:
$P(A|B) = \frac{0.2}{0.5} = 0.4$

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 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: 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: Addition Law, Mutually Exclusive Events

b) $\frac{13}{45}$
[Solution Description]
Total balls = $5 + 3 + 2 = 10$.
Probability both are red: $\frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}$.
Probability both are blue: $\frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$.
Since these are mutually exclusive events (cannot happen together), total probability is their sum:
$\frac{2}{9} + \frac{1}{15} = \frac{10}{45} + \frac{3}{45} = \frac{13}{45}$.

Key Concept: Complementary events, Independence

c) 0.35
[Solution Description]
Using De Morgan's Law and independence:
$P(M' \cap N') = P((M \cup N)') = 1 - P(M \cup N)$.
First, compute $P(M \cup N) = P(M) + P(N) - P(M \cap N) = 0.3 + 0.5 - (0.3 \times 0.5) = 0.65$.
Thus, $P(M' \cap N') = 1 - 0.65 = 0.35$.

Key Concept: Probability of exactly one event, Independent events

b) 0.5
[Solution Description]
The probability that exactly one of the two events occurs is given by $P(A \cap \overline{B}) + P(\overline{A} \cap B)$.
First, calculate $P(\overline{A}) = 1 - 0.5 = 0.5$ and $P(\overline{B}) = 1 - 0.4 = 0.6$.
Then, $P(A \cap \overline{B}) = P(A) \times P(\overline{B}) = 0.5 \times 0.6 = 0.3$.
Similarly, $P(\overline{A} \cap B) = P(\overline{A}) \times P(B) = 0.5 \times 0.4 = 0.2$.
Now add these probabilities: $0.3 + 0.2 = 0.5$.
Therefore, the probability that exactly one of the two events occurs is 0.5.

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: 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: Addition Law for Exclusive Events

b) 0.8
[Solution Description]
For mutually exclusive events, the addition law states: $P(X \cup Y) = P(X) + P(Y)$. Substituting the given values, we get $P(X \cup Y) = 0.3 + 0.5 = 0.8$.

Key Concept: Dependent Events - Marbles

a) $\frac{10}{91}$
[Solution Description]
Let event $A$: First marble is white, event $B$: Second marble is black, and event $C$: Third marble is black.
Total marbles = 8 + 6 = 14.
$P(A) = \frac{8}{14} = \frac{4}{7}$.
After drawing one white marble, remaining marbles are 7 white and 6 black.
$P(B|A) = \frac{6}{13}$.
After drawing one more black marble, remaining black marbles = 5, total marbles = 12.
$P(C|A \cap B) = \frac{5}{12}$.
The required probability is:
$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B) = \frac{4}{7} \cdot \frac{6}{13} \cdot \frac{5}{12} = \frac{120}{1092} = \frac{10}{91}$

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: Independent Events

b) $0.2$
[Solution Description]
For independent events $A$ and $B$, the probability of both occurring is:
$P(A \cap B) = P(A) \cdot P(B) = 0.4 \times 0.5 = 0.2$.

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: Addition Law for Non-Exclusive Events

c) 0.8
[Solution Description]
The events of liking football and cricket are not mutually exclusive since some students like both. Using the general addition rule: $P(F \cup C) = P(F) + P(C) - P(F \cap C)$. Given, $P(F) = \frac{60}{100} = 0.6$, $P(C) = \frac{50}{100} = 0.5$, and $P(F \cap C) = \frac{30}{100} = 0.3$. Thus, $P(F \cup C) = 0.6 + 0.5 - 0.3 = 0.8$.

Key Concept: Conditional Probability

c) 0.6667
[Solution Description]
The conditional probability $P(A | B)$ is calculated as $\frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} \approx 0.6667$.

Key Concept: Dependent Events

b) $\frac{2}{15}$
[Solution Description]
The probability of drawing the first red ball is $\frac{4}{10}$. Since the ball is not replaced, the probability of drawing a second red ball becomes $\frac{3}{9}$. The combined probability is:
$P(\text{Both Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Key Concept: Independent and Dependent Events, Multiplication Rule

b) No, because $P(\text{Football} \cap \text{Cricket}) \neq P(\text{Football}) \cdot P(\text{Cricket})$
[Solution Description]
To check if the events are independent, we verify if $P(\text{Football} \cap \text{Cricket}) = P(\text{Football}) \cdot P(\text{Cricket})$. Given: $P(\text{Football}) = 0.4$, $P(\text{Cricket}) = 0.3$, and $P(\text{Football} \cap \text{Cricket}) = 0.1$. Now, calculate $P(\text{Football}) \cdot P(\text{Cricket}) = 0.4 \times 0.3 = 0.12$. Since $0.1 \neq 0.12$, the events are dependent.

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 Dependent Events

b) $\frac{7}{11}$
[Solution Description]
Here, we need to find $P(\text{Black}_2|\text{White}_1)$. The first ball drawn is white, so it's removed from the total.
Total balls after first draw:
$7 \ (\text{black}) + 4 \ (\text{white}) = 11 \ \text{balls}$
Probability the second ball is black is:
$P(\text{Black}_2|\text{White}_1) = \frac{7}{11}$