GSEB Class 12 Maths Solutions Chapter 13 Probability Exercise 13.1

 

Question 1. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E \( \cap \) F) = 0.2, find P(E/F) and P(F/E).
Answer:
(i) \( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3} \). This calculation shows the conditional probability of E given F.
(ii) \( P(F/E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3} \). This calculates the conditional probability of F given E.
In simple words: We used the formula for conditional probability. It means finding the chance of one thing happening, given that another thing has already happened. We just plug in the known values to get the result.

Exam Tip: Remember that \( P(E/F) \) and \( P(F/E) \) are generally not equal, as they depend on the individual probabilities of E and F and their intersection.

 

Question 2. Compute P(A/B), if P(B) = 0.5 and P(A \( \cap \) B) = 0.32.
Answer:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{25} \). This calculation determines the conditional probability of A given B.
In simple words: To find the chance of A happening when B has already happened, we divide the probability of both A and B happening by the probability of B happening.

Exam Tip: Always ensure the denominator \( P(B) \) is not zero when calculating conditional probabilities, as division by zero is undefined.

 

Question 3. If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4, find
(i) P(A \( \cap \) B)
(ii) P(A/B)
(iii) P(A \( \cup \) B)
Answer:
(i) \( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
\( 0.4 = \frac{P(A \cap B)}{0.8} \)
\( \implies P(A \cap B) = 0.4 \times 0.8 = 0.32 \). This provides the probability of both events A and B occurring.

(ii) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{25} \). This represents the conditional probability of A given B.

(iii) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.8 + 0.5 - 0.32 \)
\( = 1.30 - 0.32 \)
\( = 0.98 \). This shows the probability of either A or B (or both) happening.
In simple words: We used the given information and probability formulas to find the probability of both events happening together, the chance of A happening given B, and the chance of A or B happening.

Exam Tip: Understand the relationships between \( P(A \cap B) \), \( P(A \cup B) \), and conditional probabilities like \( P(A/B) \). They are interconnected by fundamental probability rules.

 

Question 4. Evaluate P(A \( \cup \) B), if 2P(A) = P(B) = \( \frac{5}{13} \) and P(A/B) = \( \frac{2}{5} \).
Answer:
We have:
\( 2P(A) = P(B) = \frac{5}{13} \)
\( \implies P(A) = \frac{5}{26} \), and \( P(B) = \frac{5}{13} \). These are the individual probabilities of A and B.
So, \( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
\( \frac{2}{5} = \frac{P(A \cap B)}{\frac{5}{13}} \)
\( \implies P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13} \). This is the probability of both A and B occurring.
Therefore, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \)
\( = \frac{5+10-4}{26} \)
\( = \frac{11}{26} \). This is the probability of A or B happening.
In simple words: We first found the separate probabilities of A and B, then used the conditional probability formula to find the probability of both A and B happening. Finally, we used the addition rule to get the probability of A or B.

Exam Tip: When given relationships like \( 2P(A) = P(B) \), make sure to correctly deduce the individual probabilities of each event before proceeding with other calculations.

 

Question 5. If P(A) = \( \frac{6}{11} \), P(B) = \( \frac{5}{11} \) and P(A \( \cup \) B) = \( \frac{7}{11} \), find
(i) P(A \( \cap \) B)
(ii) P(A/B)
(iii) P(B/A)
Answer:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
(i) \( P(A \cap B) = P(A) + P(B) - P(A \cup B) \)
\( = \frac{6}{11} + \frac{5}{11} - \frac{7}{11} = \frac{4}{11} \). This gives the probability of both A and B happening.

(ii) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{4}{11}}{\frac{5}{11}} = \frac{4}{5} \). This is the conditional probability of A given B.

(iii) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{4}{11}}{\frac{6}{11}} = \frac{4}{6} = \frac{2}{3} \). This is the conditional probability of B given A.
In simple words: We used the addition rule for probabilities to find the chance of both A and B occurring. Then, we applied the conditional probability formula to find the chance of one event happening given another.

Exam Tip: The formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is crucial for relating union and intersection probabilities.

 

Question 6. A coin is tossed three times, where
(i) E: head on third toss, F: Heads on first two tosses
(ii) E: at least two heads, F : at most two heads
(iii) E: at most two tails F : at least one tail.
Answer:
The sample space for tossing a coin three times is \( S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \). The total number of outcomes is \( 2^3 = 8 \).

(i) E: Head occurs on third toss as
\( E = \{HHH, HTH, THH, TTH\} \). So, \( P(E) = \frac{4}{8} = \frac{1}{2} \).
F: Heads on first two tosses = \( \{HHH, HHT\} \). So, \( P(F) = \frac{2}{8} = \frac{1}{4} \).
\( E \cap F = \{HHH\} \). So, \( P(E \cap F) = \frac{1}{8} \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{8} \times 4 = \frac{1}{2} \). This is the probability of E given F.

(ii) E: At least two heads = \( \{HHT, HTH, THH, HHH\} \). So, \( P(E) = \frac{4}{8} \).
F: At most two heads = \( \{TTT, HTT, THT, TTH, HHT, HTH, THH\} \). So, \( P(F) = \frac{7}{8} \).
\( E \cap F = \{HHT, HTH, THH\} \). So, \( P(E \cap F) = \frac{3}{8} \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7} \). This is the probability of E given F.

(iii) E: At most two tails = \( \{HHT, THT, TTH, HHT, HTH, THH, HHH\} \). So, \( P(E) = \frac{7}{8} \).
F: At least one tail = \( \{THH, HTH, HHT, TTH, THT, HTT, TTT\} \). So, \( P(F) = \frac{7}{8} \).
\( E \cap F = \{HTT, THT, TTH, THH, HTH, HHT\} \). So, \( P(E \cap F) = \frac{6}{8} \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{6}{8}}{\frac{7}{8}} = \frac{6}{7} \). This is the probability of E given F.
In simple words: For each case, we listed all possible outcomes for E and F, and their shared outcomes. Then, we used these to calculate the probability of E happening when F has already occurred.

Exam Tip: Clearly listing the sample space and the elements of each event (E, F, and \( E \cap F \)) is crucial for accurately determining their probabilities in multi-toss coin problems.

 

Question 7. Two coins are tossed once, where
(i) E: tail appear on one coin, F : one coin shows head
(ii) E: no tail appears, F : no head appears.
Answer:
The sample space for tossing two coins is \( S = \{HH, HT, TH, TT\} \). The total number of outcomes is 4.

(i) E: Tail appears on one coin = \( \{TH, HT\} \). So, \( P(E) = \frac{2}{4} = \frac{1}{2} \).
F: One coin shows head = \( \{HT, TH\} \). So, \( P(F) = \frac{2}{4} = \frac{1}{2} \).
\( E \cap F = \{TH, HT\} \). So, \( P(E \cap F) = \frac{2}{4} = \frac{1}{2} \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \). This means E is certain if F happens.

(ii) E : no tail appears = \( \{HH\} \). So, \( P(E) = \frac{1}{4} \).
F : no head appears = \( \{TT\} \). So, \( P(F) = \frac{1}{4} \).
\( E \cap F = \emptyset \) (empty set). So, \( P(E \cap F) = 0 \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{\frac{1}{4}} = 0 \). This shows E cannot happen if F happens.
In simple words: For each scenario, we listed the outcomes for events E and F, then found their common outcomes. Using these, we calculated the probability of E occurring given that F has already occurred.

Exam Tip: When \( E \cap F \) is an empty set, the probability \( P(E \cap F) \) is 0, which means the conditional probability \( P(E/F) \) will also be 0, indicating the events are mutually exclusive.

 

Question 8. A dice is thrown three times, where
E: 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Answer:
A dice is thrown three times. The total number of outcomes is \( 6 \times 6 \times 6 = 216 \).
E : 4 appears on third toss
\( E = \{(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), \)
\( \hspace{1cm} (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), \)
\( \hspace{1cm} (3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4), \)
\( \hspace{1cm} (4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4), \)
\( \hspace{1cm} (5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4), \)
\( \hspace{1cm} (6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)\} \).
These are 36 cases. So, \( P(E) = \frac{36}{216} \).
F: 6 and 5 appears respectively on first two tosses
\( F = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\} \).
These are six cases. So, \( P(F) = \frac{6}{216} \).
Here, \( E \cap F = \{6, 5, 4\} \). So, \( P(E \cap F) = \frac{1}{216} \). This is the event where 6 appears on the first toss, 5 on the second, and 4 on the third.
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6} \). This gives the conditional probability of E given F.
In simple words: We first listed all possible outcomes for throwing a die three times. Then, we identified the specific outcomes for events E (4 on third toss) and F (6 then 5 on first two tosses). We found the shared outcomes and used them to calculate the probability of E happening after F has happened.

Exam Tip: When dealing with multiple dice rolls, ensure you understand how to construct the sample space and how to identify specific outcomes for defined events. Pay close attention to the exact wording, such as "appears respectively."

 

Question 9. Mother, father and son line up at random for a family picture, such that
E: Son on one end
F: Father in middle
Answer:
Let Mother be (m), Father be (f), and Son be (s).
The total number of ways they can line up is \( 3! = 3 \times 2 \times 1 = 6 \).
The sample space is \( S = \{(m, f, s), (m, s, f), (f, m, s), (f, s, m), (s, m, f), (s, f, m)\} \).
E : Son on one end : \( E = \{(s, m, f), (s, f, m), (m, f, s), (f, m, s)\} \). So, \( P(E) = \frac{4}{6} \).
F : Father in middle : \( F = \{(m, f, s), (s, f, m)\} \). So, \( P(F) = \frac{2}{6} = \frac{1}{3} \).
\( E \cap F = \{(m, f, s), (s, f, m)\} \). So, \( P(E \cap F) = \frac{2}{6} = \frac{1}{3} \). This represents the cases where the son is on one end and the father is in the middle.
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{3}}{\frac{1}{3}} = 1 \). This indicates that if the father is in the middle, the son is definitely at one end.
In simple words: We listed all the ways the family could stand in line. Then, we identified the arrangements where the son is at an end and where the father is in the middle. We found the arrangements that satisfy both conditions, and used these to find the chance of the son being at an end if the father is in the middle.

Exam Tip: For small sample spaces, writing out all possible permutations helps in accurately identifying the elements of each event and their intersection.

 

Question 10. A black and red dice are rolled:
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black dice resulted in a 5.
(b) Find the conditional probability of obtaining the sum of 8, given that the red dice resulted in a number less than 4.
Answer:
When two dice are rolled, the total number of outcomes is \( 6 \times 6 = 36 \).

(a) Let A be the event of obtaining a sum greater than 9.
\( A = \{(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\} \). So, \( P(A) = \frac{6}{36} \).
Let B be the event that the black dice resulted in a 5.
B means that the black die shows 5, and the red die can show any number.
\( B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \). So, \( P(B) = \frac{6}{36} \).
\( A \cap B = \{(5, 5), (5, 6)\} \). So, \( P(A \cap B) = \frac{2}{36} \). This is the probability of a sum greater than 9 AND the black die showing 5.
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3} \). This is the conditional probability of A given B.

(b) Let A be the event of obtaining a sum of 8.
\( A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \). So, \( P(A) = \frac{5}{36} \).
Let B be the event that the red dice resulted in a number less than 4.
This means the red die shows 1, 2, or 3, and the black die can show any number.
\( B = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)\} \). So, \( P(B) = \frac{18}{36} = \frac{1}{2} \).
\( A \cap B = \{(5, 3), (6, 2)\} \). So, \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \). This is the probability of a sum of 8 AND the red die showing less than 4.
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{18} \times 2 = \frac{1}{9} \). This is the conditional probability of A given B.
In simple words: For each part, we defined the specific events (like sum greater than 9, black die is 5) and listed their possible outcomes. Then, we found the shared outcomes and used them with the conditional probability formula to find the chance of one event happening, given the other.

Exam Tip: When dealing with two dice, carefully distinguish between "black dice" and "red dice" if they are specified, as this affects the outcomes for events like "black dice resulted in a 5."

 

Question 11. A fair dice is rolled. Consider the event E = {1, 3, 5} F = {2, 3} and G = {2, 3, 4, 5}. Find
(i) P(E/F) and P(F/E)
(ii) P(E/G), P(G/E)
(iii) P(E \( \cup \) F/G) and P(E \( \cap \) F/G)
Answer:
The sample space for a fair die roll is \( S = \{1, 2, 3, 4, 5, 6\} \). The total number of outcomes is 6.
Given: \( E = \{1, 3, 5\} \), \( F = \{2, 3\} \), \( G = \{2, 3, 4, 5\} \).
\( P(E) = \frac{3}{6} \), \( P(F) = \frac{2}{6} \), \( P(G) = \frac{4}{6} \).

(i) \( E \cap F = \{3\} \). So, \( P(E \cap F) = \frac{1}{6} \).
\( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{6}}{\frac{2}{6}} = \frac{1}{2} \). This is the probability of E given F.
\( P(F/E) = \frac{P(E \cap F)}{P(E)} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3} \). This is the probability of F given E.

(ii) \( E \cap G = \{3, 5\} \). So, \( P(E \cap G) = \frac{2}{6} \).
\( P(E/G) = \frac{P(E \cap G)}{P(G)} = \frac{\frac{2}{6}}{\frac{4}{6}} = \frac{2}{4} = \frac{1}{2} \). This is the probability of E given G.
\( P(G/E) = \frac{P(E \cap G)}{P(E)} = \frac{\frac{2}{6}}{\frac{3}{6}} = \frac{2}{3} \). This is the probability of G given E.

(iii) First, find \( E \cup F = \{1, 2, 3, 5\} \).
Then, \( (E \cup F) \cap G = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\} \).
So, \( P((E \cup F) \cap G) = \frac{3}{6} \).
\( P(E \cup F/G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{\frac{3}{6}}{\frac{4}{6}} = \frac{3}{4} \). This is the probability of E union F given G.

Next, find \( E \cap F = \{3\} \).
Then, \( (E \cap F) \cap G = \{3\} \cap \{2, 3, 4, 5\} = \{3\} \).
So, \( P((E \cap F) \cap G) = \frac{1}{6} \).
\( P(E \cap F/G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{\frac{1}{6}}{\frac{4}{6}} = \frac{1}{4} \). This is the probability of E intersection F given G.
In simple words: We used the given sets E, F, and G to find their probabilities and intersections. Then, we applied the conditional probability formula to calculate the chance of various combinations of these events happening, given another event.

Exam Tip: For conditional probability involving combinations of events (like union or intersection), calculate the probability of the combined event first, then apply the conditional probability formula using this result.

 

Question 12. Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls, given that
(i) the youngest is a girl?
(ii) at least one is a girl?
Answer:
Let G represent a girl and B represent a boy. For two children, the sample space is:
\( S = \{GG, GB, BG, BB\} \). Each outcome has a probability of \( \frac{1}{4} \).

Let A be the event that both children are girls: \( A = \{GG\} \). So, \( P(A) = \frac{1}{4} \).

(i) Let B be the event that the youngest child is a girl.
\( B = \{GG, BG\} \). So, \( P(B) = \frac{2}{4} = \frac{1}{2} \).
\( A \cap B = \{GG\} \). So, \( P(A \cap B) = \frac{1}{4} \).
The conditional probability that both are girls, given the youngest is a girl, is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \).

(ii) Let C be the event that at least one child is a girl.
\( C = \{GG, GB, BG\} \). So, \( P(C) = \frac{3}{4} \).
\( A \cap C = \{GG\} \). So, \( P(A \cap C) = \frac{1}{4} \).
The conditional probability that both are girls, given at least one is a girl, is:
\( P(A/C) = \frac{P(A \cap C)}{P(C)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \).
In simple words: We found all possible combinations for two children. Then, for each scenario, we identified the event where both are girls and the given condition (youngest is a girl, or at least one is a girl). Using these, we calculated the probability of both being girls under those specific conditions.

Exam Tip: Clearly define the sample space and the events involved. When dealing with "at least one," it's often easier to calculate the complement (none) and subtract from 1.

 

Question 13. An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy questions given that it is a multiple choice questions?
Answer:
The given data can be summarized in a table:

Let E be the event of selecting an easy question.
Let M be the event of selecting a multiple choice question.

Number of easy multiple choice questions = 500.
Total number of questions = 1400.
\( P(E \cap M) \) = Probability of selecting an easy and multiple choice question
\( = \frac{500}{1400} \).

Total number of multiple choice questions = 500 + 400 = 900.
\( P(M) \) = Probability of selecting one multiple choice question
\( = \frac{900}{1400} \).

The probability that it will be an easy question given that it is a multiple choice question is:
\( P(E/M) = \frac{P(E \cap M)}{P(M)} = \frac{\frac{500}{1400}}{\frac{900}{1400}} = \frac{500}{900} = \frac{5}{9} \).
In simple words: We first put all the question types into a table. Then, we found the chance of picking an easy multiple choice question. We also found the chance of picking any multiple choice question. Finally, we divided these to get the probability of an easy question, given that it's a multiple choice one.

Exam Tip: For problems involving multiple categories, organizing the data into a two-way table makes it much easier to identify the necessary probabilities for conditional calculations.

 

Question 14. Given that two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of the numbers on the dice is 4'.
Answer:
When two dice are thrown, the total number of exhaustive cases is \( 6 \times 6 = 36 \).
Let B be the event that the numbers appearing on the two dice are different.
The number of cases where doublets occur (same number on both dice) is 6: \( \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} \).
Number of cases when doublets do not occur = Total cases - Doublets = \( 36 - 6 = 30 \).
So, \( P(B) = \frac{30}{36} \).

Let A be the event that the sum of the numbers on the dice is 4.
The cases when the sum is 4 are \( \{(1, 3), (2, 2), (3, 1)\} \).
\( A \cap B \) denotes the event when the sum of numbers on two dice is 4 AND the numbers appearing on the dice are different.
\( A \cap B = \{(1, 3), (3, 1)\} \). So, \( P(A \cap B) = \frac{2}{36} \).

The conditional probability of A given B is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{36}}{\frac{30}{36}} = \frac{2}{30} = \frac{1}{15} \).
In simple words: We first found the total ways two different numbers can appear on two dice. Then, we listed the ways the numbers can add up to 4. We found the cases where both conditions are met and used this to calculate the probability of the sum being 4, given that the dice show different numbers.

Exam Tip: When dealing with conditions like "numbers are different," always account for the reduction in the sample space. This often involves subtracting cases with "doublets" from the total possibilities.

 

Question 15. Consider the experiment of throwing a dice, 'if a multiple of 3 comes up, throw the dice again and if any other number comes toss a coin'. Find the conditional probability of the event 'the coin shows tail', given that at least one dice shows a 3.
Answer:
Let's define the sample space based on the experiment:
If a multiple of 3 (3 or 6) comes up on the first throw, throw the dice again. Outcomes: (3,1), (3,2), ..., (3,6), (6,1), ..., (6,6). (12 outcomes)
If any other number (1, 2, 4, 5) comes up, toss a coin. Outcomes: (1,H), (1,T), (2,H), (2,T), (4,H), (4,T), (5,H), (5,T). (8 outcomes)
Total sample space size = 12 + 8 = 20.

Let E be the event 'the coin shows tail'.
\( E = \{(1,T), (2,T), (4,T), (5,T)\} \). So, \( P(E) = \frac{4}{20} = \frac{1}{5} \).

Let F be the event 'at least one die shows a 3'.
Outcomes where a multiple of 3 comes up first (dice thrown again):
First die is 3: \((3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\)
First die is 6, second die is 3: \((6,3)\)
Outcomes where a non-multiple of 3 comes up first (coin tossed):
If the first die is 3, that contradicts "non-multiple of 3", so these are not possible. However, the condition "at least one die shows a 3" means we must consider if a 3 appeared in *any* throw, including the first roll that led to a coin toss.
The question structure suggests "at least one die shows a 3" refers to the die rolls, not the coin toss outcomes.
The interpretation: "at least one die shows a 3" means the initial roll was a 3 OR the second die roll (if a multiple of 3 was rolled initially) was a 3.
\( F = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)\} \). These are 7 outcomes.
In the coin toss path, if the first die was 3, that would be a multiple of 3, leading to another die roll. So a '3' cannot appear in the first roll if a coin is tossed. This means, if a coin is tossed, there's no die showing a 3 in that path.
So the events are mutually exclusive in this setup. Let's re-evaluate "at least one die shows a 3". It seems to refer to the outcomes where a die was rolled. The coin outcomes cannot have a '3' in the die part of their description as per the rule.

Re-interpreting: The sample space is: S = {(3,1)...(3,6), (6,1)...(6,6), (1,H), (1,T), (2,H), (2,T), (4,H), (4,T), (5,H), (5,T)}.
Let A = 'coin shows tail' = \((1,T), (2,T), (4,T), (5,T)\).
Let B = 'at least one die shows a 3'.
Outcomes in B: \((3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)\).
In this interpretation, A and B are mutually exclusive. \( A \cap B = \emptyset \).
So, \( P(A \cap B) = 0 \).
And \( P(B) = \frac{7}{20} \).
\( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{\frac{7}{20}} = 0 \).
This means it's impossible for the coin to show tail if at least one die shows a 3, because if a die shows a 3, it leads to another die roll, not a coin toss.
In simple words: We analyzed the two paths of the experiment. If a die shows a 3, it always leads to throwing the die again, not tossing a coin. So, it's impossible for the coin to show a tail if a die shows a 3. Therefore, the probability is zero.

Exam Tip: Carefully map out the conditional branches of multi-stage experiments to correctly identify the sample space and the events. Ensure the conditions for each stage are mutually exclusive or inclusive as described.

 

Question 16. If P(A) = \( \frac{1}{2} \), P(B) = 0, then P(A/B) is
(a) 0
(b) \( \frac{1}{2} \)
(c) not defined
(d) 1
Answer: (c) not defined
Answer:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \).
Given \( P(B) = 0 \).
Therefore, \( P(A/B) = \frac{P(A \cap B)}{0} \). This is undefined because division by zero is not allowed.
In simple words: When the chance of event B happening is zero, we cannot calculate the conditional probability of A given B. This is because the formula requires dividing by the probability of B, which would be dividing by zero.

Exam Tip: Always check if the probability of the conditioning event (the denominator in conditional probability) is zero. If it is, the conditional probability is undefined.

 

Question 17. If A and B are events such that P(A/B) = P(B/A), then
(a) A \( \subset \) B
(b) B = A
(c) A \( \cap \) B = \( \emptyset \)
(d) P(A) = P(B)
Answer: (d) P(A) = P(B)
Answer:
Given \( P(A/B) = P(B/A) \).
Using the formula for conditional probability:
\( \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)} \).
If \( P(A \cap B) \neq 0 \), we can cancel \( P(A \cap B) \) from both sides:
\( \frac{1}{P(B)} = \frac{1}{P(A)} \).
\( \implies P(A) = P(B) \).
If \( P(A \cap B) = 0 \), then \( 0 = 0 \), which means the equality holds for any \( P(A) \) and \( P(B) \). However, the options provided offer a specific relationship. If \( P(A/B) \) and \( P(B/A) \) are both defined, then \( P(A) > 0 \) and \( P(B) > 0 \). In this case, \( P(A) = P(B) \) is the necessary condition.
In simple words: If the chance of A happening given B is the same as the chance of B happening given A, and both events can actually occur, then the overall probability of event A must be equal to the overall probability of event B.

Exam Tip: This problem demonstrates a key property of conditional probabilities. When \( P(A/B) = P(B/A) \), it often simplifies to \( P(A) = P(B) \), provided both \( P(A) \) and \( P(B) \) are non-zero.