Class 12 NCERT Maths – Chapter 13(Probability), Miscellaneous Exercise Solutions

Question 1: A and B are two events such that P(A)\ne0 . Find P(B|A) , if (i) A is a subset of B (ii) A\cap B=\phi

Solution:
Formula for Conditional Probability: P(B|A) = \frac{P(B\cap A)}{P(A)}, \quad \text{provided } P(A) \ne 0

(i) Case 1: A is a subset of B ( A \subset B )
When event A is a subset of event B (A \subset B ), it means that the occurrence of A automatically implies the occurrence of B.

In set notation, A \cap B = A
Therefore, the probability of the intersection is: P(A \cap B) = P(A)

Substituting this into the conditional probability formula: P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)}

Since P(A) \ne 0 is given, we can simplify this expression: P(B|A) = 1

(ii) Case 2: A \cap B = \phi
When the intersection of events A and B is a null set (A \cap B = \phi ), the events are said to be mutually exclusive. They have no outcomes in common.

If the events are mutually exclusive, the probability of their simultaneous occurrence (their intersection) is 0:
i.e. P(A \cap B) = P(\phi) = 0
Substituting this value into the conditional probability formula:P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)}

Since P(A) \ne 0 is given: P(B|A) = 0

Final Answer: \boxed{(i) \frac{P(B)}{P(A)} = 1, (ii) \frac{P(B)}{P(A)} = 0}

Question 2: A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.

Solution:
Let b stand for boy and g for girl.
The sample space for a couple having two children is: S = {(b, b), (b, g), (g, b), (g, g)}
Total possible outcomes n(S) = 4. We assume all outcomes are equally likely.

(i) Probability that both are males given at least one is male
Let E be the event that both children are males and F be the event that at least one of the children is male.
E = {(b, b)}
F = {(b, b), (b, g), (g, b)}
The intersection of the two events is:
E \cap F = {(b, b)}

The probabilities are:
P(F) = \frac{3}{4}

P(E \cap F) = \frac{1}{4}

Using the formula for conditional probability:
P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}

(ii) Probability that both are females given the elder is female
Let A be the event that both children are females and B be the event that the elder child is a female.
A = {(g, g)}
B = {(g, b), (g, g)} (Assuming the first coordinate represents the elder child)
The intersection of the two events is:
A \cap B = {(g, g)}

The probabilities are:
P(B) = \frac{2}{4} = \frac{1}{2}

P(A \cap B) = \frac{1}{4}

Using the formula for conditional probability:
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{2}{4} = \frac{1}{2}

Final Answer: \boxed{(i) \frac{1}{3}, (ii) \frac{1}{2}}

Question 3: Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Solution:
This problem can be solved using Bayes’ Theorem.
Let E_1 be the event that the selected person is a male and E_2 be the event that the selected person is a female.
Let A be the event that the selected person has grey hair.
Given that there are equal numbers of males and females:
P(E_1) = P(E_2) = \frac{1}{2}

The probability of having grey hair given the person is male:
P(A|E_1) = 5% = \frac{5}{100} = 0.05

The probability of having grey hair given the person is female:
P(A|E_2) = 0.25% = \frac{0.25}{100} = 0.0025

We need to find the probability that the person is male given they have grey hair, i.e., P(E_1|A).
Using Bayes’ Theorem: P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

Substituting the values:
P(E_1|A) = \frac{\frac{1}{2} \times 0.05}{(\frac{1}{2} \times 0.05) + (\frac{1}{2} \times 0.0025)}

P(E_1|A) = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525}

P(E_1|A) = \frac{500}{525} = \frac{20}{21}

Final Answer: \boxed{\frac{20}{21}}

Question 4: Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution:
This is a Bernoulli trial problem which follows the Binomial Distribution.
Let X denote the number of right-handed people in the sample.
n = 10 (Number of trials)34
p = 90% = 0.9 (Probability of success, i.e., right-handed)56
q = 1 - p = 0.1 (Probability of7 failure)8

We need to find the probability that at most 6 people are right-handed, which is P(X \le 6).
Using the property of probability:
P(X \le 6) = 1 - P(X > 6)
P(X \le 6) = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)]

The probability mass function is P(X=r) = {}^{n}C_r p^r q^{n-r}.
P(X=7) = {}^{10}C_7 (0.9)^7 (0.1)^3
P(X=8) = {}^{10}C_8 (0.9)^8 (0.1)^2
P(X=9) = {}^{10}C_9 (0.9)^9 (0.1)^1
P(X=10) = {}^{10}C_{10} (0.9)^{10} (0.1)^0

So, the expression becomes:

P(X \le 6) = 1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}

Final Answer: \boxed{1 - \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}}

Question 5: If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Solution:
A leap year has 366 days.
366 \text{ days} = 52 \text{ weeks} + 2 \text{ days}

A leap year surely has 52 Tuesdays. For it to have 53 Tuesdays, one of the remaining 2 days must be a Tuesday.

The sample space for the two consecutive remaining days is:
S = {(\text{Sun, Mon}), (\text{Mon, Tue}), (\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Sat, Sun})}

Total outcomes n(S) = 7.
Let E be the event that the leap year contains 53 Tuesdays. This happens if the pair of days includes a Tuesday.

Favourable outcomes for E are: {(\text{Mon, Tue}), (\text{Tue, Wed})}
Number of favourable outcomes n(E) = 2.

Probability P(E) = \frac{n(E)}{n(S)}

P(E) = \frac{2}{7}

Final Answer: \boxed{\frac{2}{7}}

Question 6: Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Solution:
Let E_1, E_2, E_3, E_4 be the events of selecting Box A, B, C, and D respectively.

Since the boxes are selected at random:
P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}

Let R be the event that a red marble is drawn.
The probability of drawing a red marble from each box is:

P(R|E_1) = \frac{1}{10} (Box A has 1 red out of 10)

P(R|E_2) = \frac{6}{10} (Box B has 6 red out of 10)

P(R|E_3) = \frac{8}{10} (Box C has 8 red out of 10)

P(R|E_4) = \frac{0}{10} = 0 (Box D has 0 red out of 10)

First, we calculate the total probability of drawing a red marble P(R):

P(R) = P(E_1)P(R|E_1) + P(E_2)P(R|E_2) + P(E_3)P(R|E_3) + P(E_4)P(R|E_4)

P(R) = \frac{1}{4}(\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0) = \frac{1}{4}(\frac{15}{10}) = \frac{3}{8}

Now we apply Bayes’ Theorem for each box.

(i) Probability it was from Box A:
P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(R)} = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{3}{8}} = \frac{\frac{1}{40}}{\frac{3}{8}} = \frac{1}{15}

(ii) Probability it was from Box B:
P(E_2|R) = \frac{P(E_2)P(R|E_2)}{P(R)} = \frac{\frac{1}{4} \times \frac{6}{10}}{\frac{3}{8}} = \frac{\frac{6}{40}}{\frac{3}{8}} = \frac{6}{15} = \frac{2}{5}

(iii) Probability it was from Box C:
P(E_3|R) = \frac{P(E_3)P(R|E_3)}{P(R)} = \frac{\frac{1}{4} \times \frac{8}{10}}{\frac{3}{8}} = \frac{\frac{8}{40}}{\frac{3}{8}} = \frac{8}{15}

Final Answer: \boxed{\text{Box A: } \frac{1}{15}, \text{Box B: } \frac{2}{5}, \text{Box C: } \frac{8}{15}}

Question 7: Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Solution:
Let E_1 be the event that the patient chooses Meditation and Yoga.
Let E_2 be the event that the patient chooses the Drug prescription.

Let A be the event that the patient suffers a heart attack.

Given probabilities of choosing options are equal:
P(E_1) = P(E_2) = \frac{1}{2}

The initial probability of a heart attack is 40% or 0.40.

If E_1 happens, risk reduces by 30%.
P(A|E_1) = 0.40 \times (1 - \frac{30}{100}) = 0.40 \times 0.70 = 0.28

If E_2 happens, risk reduces by 25%.
P(A|E_2) = 0.40 \times (1 - \frac{25}{100}) = 0.40 \times 0.75 = 0.30

We need to find P(E_1|A) using Bayes’ Theorem:
P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

Substituting the values:
P(E_1|A) = \frac{\frac{1}{2} \times 0.28}{(\frac{1}{2} \times 0.28) + (\frac{1}{2} \times 0.30)}

P(E_1|A) = \frac{0.14}{0.14 + 0.15} = \frac{0.14}{0.29}

P(E_1|A) = \frac{14}{29}

Final Answer: \boxed{\frac{14}{29}}

Here are the step-by-step solutions for questions 8 to 11 from the Miscellaneous Exercise on Chapter 13, formatted for your use.

Question 8: If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability \frac{1}{2}).

Solution:
A second-order determinant is given by \Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix},
where a, b, c, d \in {0, 1}.

The total number of ways to choose the 4 entries is 2^4 = 16.
Since each entry is chosen independently with probability \frac{1}{2}, each of these 16 outcomes is equally likely 1.

The value of the determinant is \Delta = ad - bc.
For the determinant to be positive, we must have ad - bc > 0.
Since the entries are only 0 or 1, the maximum value of ad or bc is 1.
Therefore, the only case where ad - bc > 0 is when ad = 1 and bc = 0.

Condition 1: ad = 1
This implies both a = 1 and d = 1. (1 way)

Condition 2: bc = 0
This implies that at least one of b or c is 0. The possible pairs (b, c) are:

(0, 0)
(0, 1)
(1, 0)(Total 3 ways)

Combining these, the favorable outcomes for the determinant are:
\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}, \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}

Total favorable outcomes = 1 \times 3 = 3.
The required probability is:
P(\text{Positive Determinant}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{16}

Final Answer: \boxed{\frac{3}{16}}

Question 9: An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A \text{ fails}) = 0.2, P(B \text{ fails alone}) = 0.15, P(A \text{ and } B \text{ fail}) = 0.15. Evaluate the following probabilities (i) P(A \text{ fails } | B \text{ has failed}) (ii) P(A \text{ fails alone})

Solution:
Let E_A be the event that subsystem A fails and E_B be the event that subsystem B fails.

We are given:
P(E_A) = 0.2
“B fails alone” means B fails but A does not. P(E_B \cap E_A') = 0.15
“A and B fail” means intersection. P(E_A \cap E_B) = 0.15

(i) Evaluate P(A \text{ fails } | B \text{ has failed})
We need to find P(E_A | E_B). The formula is:
P(E_A | E_B) = \frac{P(E_A \cap E_B)}{P(E_B)}

First, we must find P(E_B).
We know that P(E_B) = P(E_B \text{ fails alone}) + P(\text{both fail})
P(E_B) = P(E_B \cap E_A') + P(E_A \cap E_B)
P(E_B) = 0.15 + 0.15 = 0.30

Now substitute the values into the conditional probability formula:
P(E_A | E_B) = \frac{0.15}{0.30} = \frac{1}{2} = 0.5

(ii) Evaluate P(A \text{ fails alone})
“A fails alone” means A fails but B does not. This is represented by P(E_A \cap E_B').
P(E_A \cap E_B') = P(E_A) - P(E_A \cap E_B)
P(E_A \cap E_B') = 0.2 - 0.15 = 0.05

Final Answer: \boxed{\text{(i) } 0.5 \quad \text{(ii) } 0.05}

Question 10: Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution:
Let E_1 be the event that a Red ball is transferred from Bag I to Bag II.
Let E_2 be the event that a Black ball is transferred from Bag I to Bag II.

Let A be the event that the ball drawn from Bag II is Red.
Step 1: Calculate probabilities of transferring balls
Bag I has 3 Red and 4 Black balls (Total balls 7)

P(E_1) = \frac{3}{7}

P(E_2) = \frac{4}{7}

Step 2: Calculate conditional probabilities of drawing Red from Bag II
Bag II initially has 4 Red and 5 Black balls (Total balls 9).

• Case 1 (If E_1 occurs): One Red ball is added to Bag II.New composition: 4+1=5 Red, 5 Black. Total = 10.
  P(A|E_1) = \frac{5}{10} = \frac{1}{2}
• Case 2 (If E_2 occurs): One Black ball is added to Bag II.New composition: 4 Red, 5+1=6 Black. Total = 10.
  P(A|E_2) = \frac{4}{10} = \frac{2}{5}

Step 3: Apply Bayes’ Theorem
We need to find the probability that the transferred ball was black, given the drawn ball is red: P(E_2|A). 6
P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

Substitute the values:
P(E_2|A) = \frac{\frac{4}{7} \times \frac{2}{5}}{(\frac{3}{7} \times \frac{1}{2}) + (\frac{4}{7} \times \frac{2}{5})}

P(E_2|A) = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}}

To simplify, find the common denominator (LCM of 14 and 35 is 70):

Numerator: \frac{16}{70}

Denominator: \frac{15}{70} + \frac{16}{70} = \frac{31}{70}

P(E_2|A) = \frac{16}{70} \times \frac{70}{31} = \frac{16}{31}

Final Answer: \boxed{\frac{16}{31}}

Question 11: If A and B are two events such that P(A)\ne0 and P(B|A)=1, then

(A) A\subset B

(B) B\subset A

(C) B=\phi

(D) A=\phi

Solution:
We are given the conditional probability formula:
P(B|A) = \frac{P(A \cap B)}{P(A)}

According to the problem, P(B|A) = 1. 7
Substituting this into the formula:
1 = \frac{P(A \cap B)}{P(A)}
\implies P(A \cap B) = P(A)

This equation implies that the probability of the intersection of A and B is equal to the probability of A itself.
this occurs if and only if the event A is fully contained within event B.

Thus, A is a subset of B.

A \subset B

Correct Option (A).

Question 12: If P(A|B) > P(A), then which of the following is correct:

(A) P(B|A) < P(B)

(B) P(A \cap B) < P(A) \cdot P(B)

(C) P(B|A) > P(B)

(D) P(B|A) = P(B)

Solution: We are given the inequality:
P(A|B) > P(A)
Formula for conditional probability P(A|B) = \frac{P(A \cap B)}{P(B)}, substituting this into the inequality:

\frac{P(A \cap B)}{P(B)} > P(A)

Since probability P(B) is positive, we can multiply both sides by P(B):
P(A \cap B) > P(A) \cdot P(B) … (1)

Now, we need to compare this to P(B|A). The formula for P(B|A) is:
P(B|A) = \frac{P(A \cap B)}{P(A)}

P(B|A) \cdot P(A) = P(A \cap B)

Substituting in (1)
P(B|A) \cdot P(A) > P(A) \cdot P(B)
P(B|A) > P(B)

Correct Option (C).

Question 13: If A and B are any two events such that P(A) + P(B) - P(A \text{ and } B) = P(A), then

(A) P(B|A) = 1

(B) P(A|B) = 1

(C) P(B|A) = 0

(D) P(A|B) = 0

Solution: We are given the equation:
P(A) + P(B) - P(A \cap B) = P(A)

Subtracting P(A) from both sides:
P(B) - P(A \cap B) = 0
\implies P(A \cap B) = P(B)

Now we check the options which involve conditional probabilities. Let’s check P(A|B):
P(A|B) = \frac{P(A \cap B)}{P(B)}

Substitute P(A \cap B) = P(B) into the numerator:
P(A|B) = \frac{P(B)}{P(B)}

P(A|B) = 1

Correct Option (B).