RBSE Solutions Class 12 Maths Chapter 16 Probability and Probability Distribution Exercise 16.1

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.1

Question 1. If \( P(A)=\frac {7}{13}, P(B)=\frac {9}{13} \) and \( P(A \cap B) = \frac {4}{13} \), then find \( P(\frac {A}{B}) \).
Answer: We need to find the conditional probability of event A happening, given that event B has already happened. The formula for conditional probability \( P(\frac {A}{B}) \) is \( \frac{P(A \cap B)}{P(B)} \).
\( P(\frac {A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac {A}{B}) = \frac{\frac{4}{13}}{\frac{9}{13}} \)
\( P(\frac {A}{B}) = \frac{4}{13} \times \frac{13}{9} \)
\( P(\frac {A}{B}) = \frac{4}{9} \)
In simple words: To find the chance of A happening when B has already happened, we divide the chance of both A and B happening by the chance of B happening. The '13' cancels out, leaving us with 4 divided by 9.

🎯 Exam Tip: Remember the conditional probability formula \( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \) and do not confuse it with \( P(\frac{B}{A}) \), which has \( P(A) \) in the denominator.

Question 3. If \( 2P(A) = P(B) = \frac {5}{13} \) and \( P(\frac {A}{ B }) = \frac {2}{5} \), then find \( P(A \cup B) \).
Answer: We are given \( 2P(A) = \frac{5}{13} \), so \( P(A) = \frac{5}{2 \times 13} = \frac{5}{26} \). We are also given \( P(B) = \frac{5}{13} \) and \( P(\frac {A}{B}) = \frac{2}{5} \).
First, we use the conditional probability formula to find \( P(A \cap B) \):
\( P(\frac {A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( \frac{2}{5} = \frac{P(A \cap B)}{\frac{5}{13}} \)
\( P(A \cap B) = \frac{2}{5} \times \frac{5}{13} \)
\( P(A \cap B) = \frac{2}{13} \)
Now, we use the formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \)
To add and subtract these fractions, we find a common denominator, which is 26:
\( P(A \cup B) = \frac{5}{26} + \frac{5 \times 2}{13 \times 2} - \frac{2 \times 2}{13 \times 2} \)
\( P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} \)
\( P(A \cup B) = \frac{5 + 10 - 4}{26} \)
\( P(A \cup B) = \frac{11}{26} \)
In simple words: First, we found the probability of both events A and B happening together. Then, we used a formula that adds the individual probabilities of A and B, and subtracts the probability of them both happening, to find the chance of either A or B (or both) happening.

🎯 Exam Tip: When given relationships like \( 2P(A) = P(B) = k \), always solve for individual probabilities \( P(A) \) and \( P(B) \) first. Also, remember the two key formulas: conditional probability and the union of events.

Question 4. If \( P(A) = 0.6 \), \( P(B) = 0.3 \) and \( P(A \cap B) = 0.2 \), then find \( P(\frac {A}{B}) \) and \( P(\frac { B }{ A }) \).
Answer: We need to find two conditional probabilities here.
First, find \( P(\frac {A}{B}) \):
\( P(\frac {A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac {A}{B}) = \frac{0.2}{0.3} \)
\( P(\frac {A}{B}) = \frac{2}{3} \)
Next, find \( P(\frac {B}{A}) \):
\( P(\frac {B}{A}) = \frac{P(A \cap B)}{P(A)} \)
\( P(\frac {B}{A}) = \frac{0.2}{0.6} \)
\( P(\frac {B}{A}) = \frac{1}{3} \)
In simple words: We used the conditional probability formula twice. For the first part, we divided the chance of both A and B happening by the chance of B happening. For the second part, we divided the chance of both A and B happening by the chance of A happening. This shows how conditional probabilities differ depending on which event is given.

🎯 Exam Tip: Pay close attention to which event is the 'given' event, as it determines the denominator in the conditional probability formula. \( P(\frac{A}{B}) \) is not the same as \( P(\frac{B}{A}) \).

Question 5. If \( P(A) = 0.8 \), \( P(B) = 0.5 \) and \( P(\frac {B}{A}) = 0.4 \), then find:
(i) \( P(A \cap B) \)
(ii) \( P(\frac{A}{B}) \)
(iii) \( P(A \cup B) \)
Answer:
(i) We use the conditional probability formula \( P(\frac{B}{A}) = \frac{P(A \cap B)}{P(A)} \).
We are given \( P(\frac{B}{A}) = 0.4 \) and \( P(A) = 0.8 \).
\( 0.4 = \frac{P(A \cap B)}{0.8} \)
\( P(A \cap B) = 0.4 \times 0.8 \)
\( P(A \cap B) = 0.32 \)
(ii) Now we find \( P(\frac{A}{B}) \) using the formula \( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \).
We found \( P(A \cap B) = 0.32 \) in part (i), and we are given \( P(B) = 0.5 \).
\( P(\frac{A}{B}) = \frac{0.32}{0.5} \)
\( P(\frac{A}{B}) = \frac{32}{50} \)
\( P(\frac{A}{B}) = \frac{16}{25} \)
\( P(\frac{A}{B}) = 0.64 \)
(iii) Finally, we find \( P(A \cup B) \) using the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
We are given \( P(A) = 0.8 \), \( P(B) = 0.5 \), and we found \( P(A \cap B) = 0.32 \).
\( P(A \cup B) = 0.8 + 0.5 - 0.32 \)
\( P(A \cup B) = 1.3 - 0.32 \)
\( P(A \cup B) = 0.98 \)
In simple words: We first found the probability of both events A and B happening by rearranging the conditional probability formula. Then, we used this result to find the conditional probability of A given B. Lastly, we found the probability of either A or B happening using the standard union formula. This question uses all three main probability concepts together.

🎯 Exam Tip: This question is a good test of understanding the relationships between \( P(A) \), \( P(B) \), \( P(A \cap B) \), \( P(A \cup B) \), \( P(\frac{A}{B}) \), and \( P(\frac{B}{A}) \). Always write down the given values and the formulas you intend to use.

Question 6. A family has two children. What is the probability that both the children are boys, given that at least one of them is a boy?
Answer: Let's list the possible outcomes for two children. The sample space S is:
\( S = \{BB, BG, GB, GG\} \) (where B is Boy, G is Girl)
Let A be the event that at least one child is a boy.
\( A = \{BB, BG, GB\} \)
So, the number of outcomes in A is \( n(A) = 3 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{4} \).
Let B be the event that both children are boys.
\( B = \{BB\} \)
So, the number of outcomes in B is \( n(B) = 1 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{1}{4} \).
Now, let's find the intersection of A and B, which means both events A and B happen.
\( A \cap B = \{BB\} \) (Since if both are boys, at least one is a boy).
So, the number of outcomes in \( A \cap B \) is \( n(A \cap B) = 1 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{4} \).
We need to find the probability that both children are boys, given that at least one of them is a boy. This is \( P(\frac{B}{A}) \).
\( P(\frac{B}{A}) = \frac{P(A \cap B)}{P(A)} \)
\( P(\frac{B}{A}) = \frac{\frac{1}{4}}{\frac{3}{4}} \)
\( P(\frac{B}{A}) = \frac{1}{3} \)
In simple words: We first listed all ways a family can have two children. Then we picked out the cases where at least one child is a boy, and the case where both are boys. We found the chance of both happening at the same time and divided it by the chance of having at least one boy. The chance is 1 out of 3.

🎯 Exam Tip: For problems involving combinations or sequences, clearly define your sample space (S) and the events (A, B) with their respective outcomes. This makes it easier to calculate probabilities and intersections accurately.

Question 7. Two coins are tossed once. Determine \( P(\frac {A}{B}) \) where
(i) A: tail appears on one coin, B: one coin shows head
(ii) A: no tail appears, B: no head appears
Answer: The possible outcomes when two coins are tossed once are:
\( S = \{HH, HT, TH, TT\} \)
The total number of outcomes is \( n(S) = 4 \).
(i) Let A be the event "tail appears on one coin".
\( A = \{HT, TH\} \)
Let B be the event "one coin shows head".
\( B = \{HT, TH\} \)
The intersection of A and B is:
\( A \cap B = \{HT, TH\} \)
So, \( n(A \cap B) = 2 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{4} = \frac{1}{2} \).
The probability of B is \( P(B) = \frac{n(B)}{n(S)} = \frac{2}{4} = \frac{1}{2} \).
Now, we find \( P(\frac{A}{B}) \):
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{\frac{1}{2}}{\frac{1}{2}} \)
\( P(\frac{A}{B}) = 1 \)
(ii) Let A be the event "no tail appears". This means both coins show heads.
\( A = \{HH\} \)
Let B be the event "no head appears". This means both coins show tails.
\( B = \{TT\} \)
The intersection of A and B is:
\( A \cap B = \emptyset \) (an empty set, as it's impossible for both conditions to be true at the same time)
So, \( n(A \cap B) = 0 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{0}{4} = 0 \).
The probability of B is \( P(B) = \frac{n(B)}{n(S)} = \frac{1}{4} \).
Now, we find \( P(\frac{A}{B}) \):
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{0}{\frac{1}{4}} \)
\( P(\frac{A}{B}) = 0 \)
In simple words: For the first part, we found that if one coin shows a head, it is certain that a tail also appears on one coin, so the probability is 1. For the second part, if no head appears, it's impossible for no tail to appear at the same time, so the probability is 0. This shows how conditional probability changes based on the given information.

🎯 Exam Tip: Always define your sample space clearly. When events are mutually exclusive (like in part ii), their intersection is an empty set, and its probability is 0.

Question 8. Mother, father and son line up at random for a family picture. Events related to it are defined as events A and B. Find \( P(\frac {A}{B}) \) if
A: Son on one end
B: Father in middle
Answer: Let Mother be M, Father be F, and Son be S.
The total number of ways they can stand in a line is the number of permutations of 3 distinct people, which is \( 3! \).
Total number of arrangements \( n(S) = 3 \times 2 \times 1 = 6 \).
The sample space S is:
\( S = \{MFS, MSF, FMS, FSM, SMF, SFM\} \)
Let A be the event that the son is on one end.
\( A = \{SMF, SFM, FMS, MFS\} \) (Son at the beginning or at the end).
So, \( n(A) = 4 \).
The probability of A is \( P(A) = \frac{n(A)}{n(S)} = \frac{4}{6} = \frac{2}{3} \).
Let B be the event that the father is in the middle.
\( B = \{MF S, SF M\} \)
So, \( n(B) = 2 \).
The probability of B is \( P(B) = \frac{n(B)}{n(S)} = \frac{2}{6} = \frac{1}{3} \).
Now, find the intersection of A and B, where the son is on one end AND the father is in the middle.
\( A \cap B = \{MFS, SFM\} \)
So, \( n(A \cap B) = 2 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{6} = \frac{1}{3} \).
We need to find \( P(\frac{A}{B}) \), the probability that the son is on one end, given the father is in the middle.
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{\frac{1}{3}}{\frac{1}{3}} \)
\( P(\frac{A}{B}) = 1 \)
In simple words: We listed all the ways the family could stand. Then, we found the chance that the son is at an end and the chance that the father is in the middle. Because all arrangements where the father is in the middle also have the son on an end, the probability of the son being at an end given the father is in the middle is 1, meaning it's certain.

🎯 Exam Tip: When dealing with permutations, listing out the sample space clearly can prevent errors. If event B is a subset of event A, then \( P(\frac{A}{B}) \) will always be 1.

Question 9. A fair die is rolled. Consider events A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}, find :
(i) \( P(\frac {A}{B}) \) and \( P(\frac {B}{A}) \)
(ii) \( P(\frac {A}{C}) \) and \( P(\frac {C}{A}) \)
(iii) \( P\left(\frac {(A \cup B)}{C}\right) \) and \( P\left(\frac {(A \cap B)}{C}\right) \)
Answer: The sample space for rolling a fair die is \( S = \{1, 2, 3, 4, 5, 6\} \). So, \( n(S) = 6 \).
Given events:
\( A = \{1, 3, 5\} \implies n(A) = 3 \implies P(A) = \frac{3}{6} = \frac{1}{2} \)
\( B = \{2, 3\} \implies n(B) = 2 \implies P(B) = \frac{2}{6} = \frac{1}{3} \)
\( C = \{2, 3, 4, 5\} \implies n(C) = 4 \implies P(C) = \frac{4}{6} = \frac{2}{3} \)

(i) Find \( P(\frac {A}{B}) \) and \( P(\frac {B}{A}) \):
First, find \( A \cap B = \{1, 3, 5\} \cap \{2, 3\} = \{3\} \). So, \( n(A \cap B) = 1 \).
\( P(A \cap B) = \frac{1}{6} \).
\( P(\frac {A}{B}) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2} \)
\( P(\frac {B}{A}) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{6} = \frac{1}{3} \)

(ii) Find \( P(\frac {A}{C}) \) and \( P(\frac {C}{A}) \):
First, find \( A \cap C = \{1, 3, 5\} \cap \{2, 3, 4, 5\} = \{3, 5\} \). So, \( n(A \cap C) = 2 \).
\( P(A \cap C) = \frac{2}{6} = \frac{1}{3} \).
\( P(\frac {A}{C}) = \frac{P(A \cap C)}{P(C)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} \)
\( P(\frac {C}{A}) = \frac{P(A \cap C)}{P(A)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3} \)

(iii) Find \( P\left(\frac {(A \cup B)}{C}\right) \) and \( P\left(\frac {(A \cap B)}{C}\right) \):
First, find \( A \cup B = \{1, 3, 5\} \cup \{2, 3\} = \{1, 2, 3, 5\} \).
Now, find \( (A \cup B) \cap C = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\} \). So, \( n((A \cup B) \cap C) = 3 \).
\( P((A \cup B) \cap C) = \frac{3}{6} = \frac{1}{2} \).
\( P\left(\frac {(A \cup B)}{C}\right) = \frac{P((A \cup B) \cap C)}{P(C)} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \)
Next, we already found \( A \cap B = \{3\} \).
Now, find \( (A \cap B) \cap C = \{3\} \cap \{2, 3, 4, 5\} = \{3\} \). So, \( n((A \cap B) \cap C) = 1 \).
\( P((A \cap B) \cap C) = \frac{1}{6} \).
\( P\left(\frac {(A \cap B)}{C}\right) = \frac{P((A \cap B) \cap C)}{P(C)} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4} \)
In simple words: This question explores conditional probability with three different events. We calculated the basic probabilities for each event and their intersections. Then, using the conditional probability formula, we found the likelihood of one event happening given another. For part (iii), we first found the union and intersection of events A and B, then found their intersection with C, and finally applied the conditional probability rule.

🎯 Exam Tip: When dealing with multiple events, clearly list the elements in each set (A, B, C) and their intersections/unions. Calculate their probabilities before applying conditional probability formulas to avoid mistakes.

Question 10. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.
Answer: When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \).
Let E be the event that the two numbers appearing on the dice are different.
The total outcomes where the numbers are the same are \(\{ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \} \). There are 6 such outcomes.
So, the number of outcomes where the numbers are different is \( n(E) = 36 - 6 = 30 \).
The probability of E is \( P(E) = \frac{30}{36} = \frac{5}{6} \).
Let F be the event that the sum of the numbers on the dice is 4.
The outcomes for event F are \(\{ (1,3), (2,2), (3,1) \} \). So, \( n(F) = 3 \).
The probability of F is \( P(F) = \frac{3}{36} = \frac{1}{12} \).
We need to find the intersection of E and F, i.e., the sum is 4 AND the numbers are different.
\( E \cap F = \{ (1,3), (3,1) \} \) (We exclude (2,2) because the numbers must be different).
So, \( n(E \cap F) = 2 \).
The probability of \( E \cap F \) is \( P(E \cap F) = \frac{2}{36} = \frac{1}{18} \).
We want to find the probability of F given E, which is \( P(\frac{F}{E}) \).
\( P(\frac{F}{E}) = \frac{P(E \cap F)}{P(E)} \)
\( P(\frac{F}{E}) = \frac{\frac{1}{18}}{\frac{5}{6}} \)
\( P(\frac{F}{E}) = \frac{1}{18} \times \frac{6}{5} \)
\( P(\frac{F}{E}) = \frac{6}{90} \)
\( P(\frac{F}{E}) = \frac{1}{15} \)
In simple words: First, we figured out all the ways two dice could land with different numbers. Then, we looked for the ways the numbers could add up to 4 while also being different. Finally, we divided the chance of getting a sum of 4 with different numbers by the chance of getting different numbers, to find our answer.

🎯 Exam Tip: When a condition is given (like "numbers are different"), always adjust your sample space or the intersection event to reflect this condition. Listing favorable outcomes clearly helps avoid errors.

Question 11. Ten cards numbered 1 to 10 are placed in a box, mixed up throughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Answer: The sample space consists of cards numbered 1 to 10.
\( S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \). So, \( n(S) = 10 \).
Let A be the event that the drawn card is an even number.
\( A = \{2, 4, 6, 8, 10\} \). So, \( n(A) = 5 \).
The probability of A is \( P(A) = \frac{5}{10} = \frac{1}{2} \).
Let B be the event that the number on the drawn card is more than 3.
\( B = \{4, 5, 6, 7, 8, 9, 10\} \). So, \( n(B) = 7 \).
The probability of B is \( P(B) = \frac{7}{10} \).
We need to find the intersection of A and B, which means the card is both even AND the number is more than 3.
\( A \cap B = \{4, 6, 8, 10\} \). So, \( n(A \cap B) = 4 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{4}{10} \).
We are asked to find the probability that the card is an even number, given that it is more than 3. This is \( P(\frac{A}{B}) \).
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{\frac{4}{10}}{\frac{7}{10}} \)
\( P(\frac{A}{B}) = \frac{4}{7} \)
In simple words: We know the card drawn is greater than 3. So, we only consider numbers from 4 to 10. Out of these numbers, we count how many are even. Then, we divide this count by the total numbers greater than 3. This gives us the chance of picking an even card when we already know it's a number higher than 3.

🎯 Exam Tip: Clearly define your events and their respective elements. When dealing with conditional probability, make sure your numerator is the probability of the intersection, and your denominator is the probability of the given event.

Question 12. In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl.
Answer: Total number of students = 1000.
Number of girls = 430.
Number of girls in Class XII = 10% of 430 = \( 0.10 \times 430 = 43 \).
Let A be the event that a student chosen randomly studies in Class XII.
Let B be the event that the chosen student is a girl.
We are asked to find \( P(\frac{A}{B}) \), which is the probability that a student studies in Class XII given that the student is a girl.
This means we are looking for the probability of A happening, restricted only to the group of girls.
The "given" event is B (the student is a girl). So our reduced sample space is the number of girls, which is 430.
The number of outcomes where A happens within this reduced sample space (girls who study in Class XII) is 43.
So, \( P(\frac{A}{B}) = \frac{\text{Number of girls in Class XII}}{\text{Total number of girls}} \)
\( P(\frac{A}{B}) = \frac{43}{430} \)
\( P(\frac{A}{B}) = \frac{1}{10} \)
\( P(\frac{A}{B}) = 0.1 \)
Alternatively, using probabilities:
\( P(B) = \frac{\text{Number of girls}}{\text{Total students}} = \frac{430}{1000} = 0.43 \)
\( P(A \cap B) = \frac{\text{Number of girls in Class XII}}{\text{Total students}} = \frac{43}{1000} = 0.043 \)
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} = \frac{0.043}{0.43} = \frac{43}{430} = \frac{1}{10} = 0.1 \)
In simple words: We know the student is a girl, so we only care about the girls. We found that 10% of the girls are in Class XII. So, the chance of a chosen girl being in Class XII is exactly that 10%, or 0.1.

🎯 Exam Tip: For conditional probability problems, you can often directly use the number of favorable outcomes within the reduced sample space (the 'given' condition) rather than calculating full probabilities first, simplifying the steps.

Question 13. A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once ?
Answer: When a die is thrown twice, the total number of possible outcomes is \( 6 \times 6 = 36 \).
Let B be the event that the sum of the numbers is 6.
The outcomes for B are: \( \{(1,5), (2,4), (3,3), (4,2), (5,1)\} \).
So, \( n(B) = 5 \).
The probability of B is \( P(B) = \frac{5}{36} \).
Let A be the event that the number 4 has appeared at least once.
The outcomes for A are: \( \{(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)\} \).
So, \( n(A) = 11 \).
Now, we need to find the intersection of A and B, i.e., the sum is 6 AND the number 4 appeared at least once.
\( A \cap B = \{(2,4), (4,2)\} \).
So, \( n(A \cap B) = 2 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{2}{36} \).
We need to find the probability that the number 4 has appeared at least once, given that the sum is 6. This is \( P(\frac{A}{B}) \).
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{\frac{2}{36}}{\frac{5}{36}} \)
\( P(\frac{A}{B}) = \frac{2}{5} \)
In simple words: We first listed all pairs of dice rolls that add up to 6. Then, from these pairs, we picked out only those where the number 4 showed up at least once. The answer is the count of these selected pairs divided by the total count of pairs that add up to 6.

🎯 Exam Tip: For dice roll problems, drawing a 6x6 grid can help visualize all possible outcomes and easily identify events and their intersections. Be careful to correctly identify "at least once" outcomes.

Question 14. Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'?
Answer: Let's first define the sample space and its probabilities based on the experiment.
If the first toss is Head (H) (probability \( \frac{1}{2} \)), toss again:
Outcomes: HH, HT (each with probability \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \))
If the first toss is Tail (T) (probability \( \frac{1}{2} \)), throw a die:
Outcomes: T1, T2, T3, T4, T5, T6 (each with probability \( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \))
The complete sample space S with probabilities:
\( S = \{HH(\frac{1}{4}), HT(\frac{1}{4}), T1(\frac{1}{12}), T2(\frac{1}{12}), T3(\frac{1}{12}), T4(\frac{1}{12}), T5(\frac{1}{12}), T6(\frac{1}{12})\} \)
Let A be the event that "the die shows a number greater than 4". This event can only happen if the first toss was a Tail.
\( A = \{T5, T6\} \)
\( P(A) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \)
Let B be the event that "there is at least one tail".
\( B = \{HT, T1, T2, T3, T4, T5, T6\} \)
\( P(B) = P(HT) + P(T1) + P(T2) + P(T3) + P(T4) + P(T5) + P(T6) \)
\( P(B) = \frac{1}{4} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} \)
\( P(B) = \frac{1}{4} + \frac{6}{12} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \)
Now, find the intersection of A and B, i.e., the die shows a number greater than 4 AND there is at least one tail.
\( A \cap B = \{T5, T6\} \) (These outcomes satisfy both conditions).
\( P(A \cap B) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \)
We need to find \( P(\frac{A}{B}) \), which is the probability of the die showing a number greater than 4, given that there is at least one tail.
\( P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)} \)
\( P(\frac{A}{B}) = \frac{\frac{1}{6}}{\frac{3}{4}} \)
\( P(\frac{A}{B}) = \frac{1}{6} \times \frac{4}{3} \)
\( P(\frac{A}{B}) = \frac{4}{18} \)
\( P(\frac{A}{B}) = \frac{2}{9} \)
In simple words: We first wrote down all possible results and their chances for the mixed coin and die experiment. Then, we found the chance of the die showing a number bigger than 4. We also found the chance of getting at least one tail. Finally, we divided the chance of both of these things happening by the chance of getting at least one tail, to get our answer.

🎯 Exam Tip: For multi-stage experiments, it's crucial to correctly calculate the probability of each final outcome by multiplying the probabilities of the stages. Clearly listing the sample space with probabilities helps define events and their intersections.