NCERT Solutions Class-12-Chapter-13-Probability
Miscellaneous Exercise on Chapter 13
Note: This miscellaneous exercise covers various concepts from the chapter:
1. Conditional Probability: \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
2. Bayes' Theorem: Calculating reverse probabilities based on new evidence.
3. Bernoulli Trials: \( P(X=r) = \binom{n}{r} p^r q^{n-r} \).
4. Independent Events: \( P(A \cap B) = P(A) \cdot P(B) \).
Q1
A and B are two events such that \( P(A) \neq 0 \). Find \( P(B|A) \), if
(i) A is a subset of B
(ii) \( A \cap B = \phi \)
(i) A is a subset of B
(ii) \( A \cap B = \phi \)
Solution:
We know that \( P(B|A) = \frac{P(A \cap B)}{P(A)} \).
(i) If \( A \subset B \):
Then \( A \cap B = A \).
Therefore, \( P(B|A) = \frac{P(A)}{P(A)} = 1 \).
(ii) If \( A \cap B = \phi \):
Then \( P(A \cap B) = 0 \).
Therefore, \( P(B|A) = \frac{0}{P(A)} = 0 \).
Q2
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.
(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:
Sample space \( S = \{MM, MF, FM, FF\} \). Total outcomes = 4.
(i) Let E = both are males = \( \{MM\} \).
Let F = at least one is male = \( \{MM, MF, FM\} \).
\( E \cap F = \{MM\} \).
\( P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{3/4} = \frac{1}{3} \).
(ii) Let A = both are females = \( \{FF\} \).
Let B = elder child is female = \( \{FM, FF\} \).
\( A \cap B = \{FF\} \).
\( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{2/4} = \frac{1}{2} \).
Q3
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:
Let \( E_1 \): Person is Male, \( E_2 \): Person is Female.
Since numbers are equal, \( P(E_1) = P(E_2) = 0.5 \).
Let A: Person has grey hair.
\( P(A|E_1) = 5\% = 0.05 \).
\( P(A|E_2) = 0.25\% = 0.0025 \).
Using Bayes' Theorem for \( P(E_1|A) \):
\( P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( = \frac{0.5 \times 0.05}{0.5 \times 0.05 + 0.5 \times 0.0025} = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525} \)
\( = \frac{500}{525} = \frac{20}{21} \).
Q4
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. Let X denote the number of right-handed people.
\( n = 10 \), \( p = 0.9 \), \( q = 0.1 \).
We need \( P(X \le 6) = 1 - P(X > 6) \)
\( = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)] \).
Using formula \( P(X=r) = \binom{n}{r} p^r q^{n-r} \):
\( = 1 - [\binom{10}{7}(0.9)^7(0.1)^3 + \binom{10}{8}(0.9)^8(0.1)^2 + \binom{10}{9}(0.9)^9(0.1)^1 + \binom{10}{10}(0.9)^{10}] \)
\( = 1 - \sum_{r=7}^{10} \binom{10}{r} (0.9)^r (0.1)^{10-r} \).
Q5
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 = 52 weeks + 2 extra days.
The sample space for the 2 extra days is:
\( S = \{(Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)\} \).
Total outcomes = 7.
Outcomes containing a Tuesday: \( (Mon, Tue), (Tue, Wed) \). (2 outcomes).
Probability = \( \frac{2}{7} \).
Q6
Suppose we have four boxes A, B, C and D containing coloured marbles as given below:Box Marble colour Red White Black A 1 6 3 B 6 2 2 C 8 1 1 D 0 6 4
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?
| Box | Marble colour | ||
|---|---|---|---|
| Red | White | Black | |
| A | 1 | 6 | 3 |
| B | 6 | 2 | 2 |
| C | 8 | 1 | 1 |
| D | 0 | 6 | 4 |
Solution:
Let \( E_A, E_B, E_C, E_D \) be events of selecting boxes. \( P(E) = \frac{1}{4} \) for each.
Let R be the event of drawing a Red marble.
\( P(R|A) = \frac{1}{10} \), \( P(R|B) = \frac{6}{10} \), \( P(R|C) = \frac{8}{10} \), \( P(R|D) = 0 \).
Total Probability \( P(R) = \frac{1}{4}(\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0) = \frac{15}{40} \).
(i) From Box A:
\( P(A|R) = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{15}{40}} = \frac{1/40}{15/40} = \frac{1}{15} \).
(ii) From Box B:
\( P(B|R) = \frac{\frac{1}{4} \times \frac{6}{10}}{\frac{15}{40}} = \frac{6/40}{15/40} = \frac{6}{15} = \frac{2}{5} \).
(iii) From Box C:
\( P(C|R) = \frac{\frac{1}{4} \times \frac{8}{10}}{\frac{15}{40}} = \frac{8/40}{15/40} = \frac{8}{15} \).
Q7
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:
Original risk = 0.40.
Let \( E_1 \): Meditation (Risk reduced by 30% \( \Rightarrow \) New risk = 70% of 0.40 = 0.28).
Let \( E_2 \): Drug (Risk reduced by 25% \( \Rightarrow \) New risk = 75% of 0.40 = 0.30).
\( P(E_1) = P(E_2) = 0.5 \).
Let A: Patient suffers Heart Attack.
\( P(A|E_1) = 0.28 \), \( P(A|E_2) = 0.30 \).
Using Bayes' Theorem for \( P(E_1|A) \):
\( P(E_1|A) = \frac{0.5 \times 0.28}{0.5 \times 0.28 + 0.5 \times 0.30} = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58} = \frac{28}{58} = \frac{14}{29} \).
Q8
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:
Total possible determinants = \( 2^4 = 16 \).
Determinant \( \Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \).
For \( \Delta > 0 \) (since entries are 0 or 1, max value is 1), we must have \( ad - bc = 1 \).
This implies \( ad = 1 \) and \( bc = 0 \).
- \( ad = 1 \implies a=1, d=1 \) (1 case).
- \( bc = 0 \implies (b=0, c=0), (b=1, c=0), (b=0, c=1) \) (3 cases).
Total favorable cases = \( 1 \times 3 = 3 \).
Probability = \( \frac{3}{16} \).
Q9
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 fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities
(i) P(A fails|B has failed) (ii) P(A fails alone)
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities
(i) P(A fails|B has failed) (ii) P(A fails alone)
Solution:
Let \( E_A \): A fails, \( E_B \): B fails.
\( P(E_A) = 0.2 \).
\( P(E_A \cap E_B) = 0.15 \).
\( P(\text{B fails alone}) = P(E_B \cap E_A') = P(E_B) - P(E_A \cap E_B) = 0.15 \).
\( \Rightarrow P(E_B) - 0.15 = 0.15 \Rightarrow P(E_B) = 0.30 \).
(i) \( P(E_A | E_B) = \frac{P(E_A \cap E_B)}{P(E_B)} = \frac{0.15}{0.30} = 0.5 \).
(ii) P(A fails alone) = \( P(E_A \cap E_B') = P(E_A) - P(E_A \cap E_B) \)
\( = 0.2 - 0.15 = 0.05 \).
Q10
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:
\( E_1 \): Transferred Red (3/7), \( E_2 \): Transferred Black (4/7).
A: Drawn Red from Bag II.
If \( E_1 \): Bag II has 5 Red, 5 Black (Total 10). \( P(A|E_1) = \frac{5}{10} = \frac{1}{2} \).
If \( E_2 \): Bag II has 4 Red, 6 Black (Total 10). \( P(A|E_2) = \frac{4}{10} = \frac{2}{5} \).
Required \( P(E_2|A) \):
\( = \frac{\frac{4}{7} \times \frac{4}{10}}{\frac{3}{7} \times \frac{5}{10} + \frac{4}{7} \times \frac{4}{10}} = \frac{16}{15 + 16} = \frac{16}{31} \).
Q11
If A and B are two events such that \( P(A) \neq 0 \) and \( P(B | A) = 1 \), then
(A) \( A \subset B \) (B) \( B \subset A \) (C) \( B = \phi \) (D) \( A = \phi \)
(A) \( A \subset B \) (B) \( B \subset A \) (C) \( B = \phi \) (D) \( A = \phi \)
Answer: (A)
Solution:
\( P(B|A) = \frac{P(A \cap B)}{P(A)} = 1 \).
\( \Rightarrow P(A \cap B) = P(A) \).
This implies the intersection of A and B is A itself, which means A is completely inside B.
So, \( A \subset B \).
Q12
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) \( P(B|A) = P(B) \)
(A) \( P(B|A) < P(B) \)
(B) \( P(A \cap B) < P(A) \cdot P(B) \)
(C) \( P(B|A) > P(B) \)
(D) \( P(B|A) = P(B) \)
Answer: (C)
Solution:
\( \frac{P(A \cap B)}{P(B)} > P(A) \)
\( \Rightarrow P(A \cap B) > P(A) \cdot P(B) \)
\( \Rightarrow \frac{P(A \cap B)}{P(A)} > P(B) \)
\( \Rightarrow P(B|A) > P(B) \).
Q13
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) \( P(A|B) = 0 \)
(A) \( P(B|A) = 1 \)
(B) \( P(A|B) = 1 \)
(C) \( P(B|A) = 0 \)
(D) \( P(A|B) = 0 \)
Answer: (B)
Solution:
\( P(A) + P(B) - P(A \cap B) = P(A) \).
\( \Rightarrow P(B) - P(A \cap B) = 0 \).
\( \Rightarrow P(A \cap B) = P(B) \).
Now, \( P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1 \).