Miscellaneous Exercise Class 12 Maths Probability

NCERT Solutions Class 12 Maths Chapter 13 Miscellaneous | LearnCBSEHub

💡 Chapter Review

This exercise combines all key concepts from the chapter:

Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$
Bayes’ Theorem: Calculating reverse probabilities.
Total Probability: Summing probabilities across mutually exclusive cases.

Question 01

Find $P(B|A)$ if (i) $A \subset B$, (ii) $A \cap B = \phi$.

(i) If $A \subset B$

Since $A \subset B$, the intersection $A \cap B = A$.
$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1$.

(ii) If $A \cap B = \phi$

Since $A \cap B = \phi$, $P(A \cap B) = 0$.
$P(B|A) = \frac{0}{P(A)} = 0$.

Question 02

Two children. (i) Prob(Both Male | At least one Male). (ii) Prob(Both Female | Elder is Female).

Sample Space $S = \{MM, MF, FM, FF\}$ (M=Male, F=Female, order matters).

(i) Let $E$: Both Male ($MM$), $F$: At least one Male ($MM, MF, FM$).
$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{3/4} = \frac{1}{3}$.

(ii) Let $A$: Both Female ($FF$), $B$: Elder is Female ($FF, FM$).
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{2/4} = \frac{1}{2}$.

Question 03

5% Men, 0.25% Women have grey hair. Person selected has grey hair. Prob(Person is Male)?

Let $E_1$: Person is Male, $E_2$: Person is Female. $P(E_1) = P(E_2) = 0.5$.
Let $A$: Person has Grey Hair.
$P(A|E_1) = 5\% = 0.05$.
$P(A|E_2) = 0.25\% = 0.0025$.
By Bayes’ Theorem:
$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
$= \frac{0.5 \times 0.05}{0.5 \times 0.05 + 0.5 \times 0.0025} = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525}$
$= \frac{500}{525} = \frac{20}{21}$.

Answer: 20/21

Question 04

90% Right-handed. Prob(At most 6 out of 10 are right-handed).

Bernoulli Trials: $n=10, p=0.9, q=0.1$.
Let $X$ be the number of right-handed people.
We need $P(X \le 6)$. Using the complement rule:
$P(X \le 6) = 1 – P(X > 6) = 1 – [P(X=7) + P(X=8) + P(X=9) + P(X=10)]$.
$= 1 – \sum_{r=7}^{10} {}^{10}C_r (0.9)^r (0.1)^{10-r}$.

Note: This uses Binomial Distribution.

Question 05

Prob(Leap year has 53 Tuesdays).

A leap year has 366 days = 52 weeks + 2 extra days.
The 2 extra days can be: (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon).
Total outcomes = 7.
Outcomes with Tuesday = (Mon, Tue) and (Tue, Wed) = 2 outcomes.
Probability = $2/7$.

Answer: 2/7

Question 06

4 Boxes (A, B, C, D) with marbles. Red marble drawn. Find Prob(Box A), Prob(Box B), Prob(Box C).

Step 1: Calculate Probabilities

$P(A)=P(B)=P(C)=P(D) = 1/4$.
Let $R$ be event ‘Red marble’.
Box A (1R, 6W, 3B, Total 10): $P(R|A) = 1/10$.
Box B (6R, 2W, 2B, Total 10): $P(R|B) = 6/10$.
Box C (8R, 1W, 1B, Total 10): $P(R|C) = 8/10$.
Box D (0R, 6W, 4B, Total 10): $P(R|D) = 0$.

Total Prob $P(R) = \frac{1}{4}(\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0) = \frac{15}{40}$.

Step 2: Apply Bayes’ Theorem

$P(A|R) = \frac{(1/4)(1/10)}{15/40} = \frac{1/40}{15/40} = \frac{1}{15}$.
$P(B|R) = \frac{(1/4)(6/10)}{15/40} = \frac{6/40}{15/40} = \frac{6}{15} = \frac{2}{5}$.
$P(C|R) = \frac{(1/4)(8/10)}{15/40} = \frac{8/40}{15/40} = \frac{8}{15}$.

Question 07

Heart Attack Risk. Yoga reduces risk by 30%, Drug by 25%. Patient has attack. Prob(Yoga)?

Base Risk = 40%.
Let $E_1$: Yoga, $E_2$: Drug. $P(E_1)=P(E_2)=0.5$.
Let $A$: Heart Attack.
If Yoga: Risk reduces by 30% of 40% $\to$ New Risk = $40 \times 0.7 = 28\% = 0.28$.
So, $P(A|E_1) = 0.28$.
If Drug: Risk reduces by 25% of 40% $\to$ New Risk = $40 \times 0.75 = 30\% = 0.30$.
So, $P(A|E_2) = 0.30$.
$P(E_1|A) = \frac{0.5 \times 0.28}{0.5 \times 0.28 + 0.5 \times 0.30} = \frac{0.14}{0.14 + 0.15} = \frac{0.14}{0.29} = \frac{14}{29}$.

Answer: 14/29

Question 08

Determinant of 2×2 matrix (entries 0 or 1). Prob(Determinant > 0).

Total matrices = $2^4 = 16$.
Det $\Delta = ad – bc$. Entries $\in \{0, 1\}$.
For $\Delta > 0$, we need $ad – bc = 1$ (since max possible value is 1).
This requires $ad=1$ AND $bc=0$.
$ad=1 \implies a=1, d=1$ (1 case).
$bc=0 \implies (b,c)$ can be $(0,0), (0,1), (1,0)$ (3 cases).
Total favorable cases = $1 \times 3 = 3$.
Probability = $3/16$.

Answer: 3/16

Question 09

Given $P(A)=0.2$, $P(B \text{ only}) = 0.15$, $P(A \cap B)=0.15$.

$P(B) = P(B \text{ only}) + P(A \cap B) = 0.15 + 0.15 = 0.3$.
(i) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.3} = 0.5$.
(ii) $P(A \text{ fails alone}) = P(A) – P(A \cap B) = 0.2 – 0.15 = 0.05$.

Question 10

Bag I (3R, 4B), Bag II (4R, 5B). Ball transferred I to II. Red drawn from II. Prob(Transferred Black).

Let $E_1$: Transferred Red ($P(E_1)=3/7$). $E_2$: Transferred Black ($P(E_2)=4/7$).
A: Drawing Red from Bag II.
If $E_1$: Bag II has $4+1=5$ Red, 5 Black (Total 10). $P(A|E_1) = 5/10 = 1/2$.
If $E_2$: Bag II has 4 Red, $5+1=6$ Black (Total 10). $P(A|E_2) = 4/10 = 2/5$.
$P(E_2|A) = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{8/35}{3/14 + 8/35}$
$= \frac{8/35}{15/70 + 16/70} = \frac{16/70}{31/70} = \frac{16}{31}$.

Answer: 16/31

Questions 11 — 13

Multiple Choice Questions.

11. If $P(B|A)=1$, then:

$\frac{P(A \cap B)}{P(A)} = 1 \implies P(A \cap B) = P(A)$.
This implies all outcomes of A are also in B. Thus, $A \subset B$.

Correct Option: (A)

12. If $P(A|B) > P(A)$, then:

$\frac{P(A \cap B)}{P(B)} > P(A) \implies P(A \cap B) > P(A)P(B)$.
Divide by $P(A)$: $\frac{P(A \cap B)}{P(A)} > P(B) \implies P(B|A) > P(B)$.

Correct Option: (C)

13. If $P(A)+P(B)-P(A \cap B) = P(A)$:

$P(B) – P(A \cap B) = 0 \implies P(A \cap B) = P(B)$.
Then $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)}$ (Not necessarily 1).
But $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$.

Correct Option: (B)