Find complete, step-by-step solutions for NCERT Class 12 Maths Exercise 13.3 (Probability). Master problems using the Theorem of Total Probability and Bayes’ Theorem

This exercise primarily involves the Theorem of Total Probability and Bayes’ Theorem.

Ex. 13.1

Ex. 13.2

Ex. 13.3

Ch 13 Misc.

Theorem of Total Probability

$P(E) = P(E|E_1)P(E_1) + P(E|E_2)P(E_2) + \dots + P(E|E_n)P(E_n)$

Bayes’ Theorem

$$P(E_i|E) = \frac{P(E|E_i)P(E_i)}{\sum_{j=1}^{n} P(E|E_j)P(E_j)}$$

1. Second Ball is Red (Total Probability)

Let $R_1$ and $B_1$ be the events that the first ball drawn is Red or Black, respectively.

Let $R_2$ be the event that the second ball drawn is Red.

Initial composition: 5 Red (R), 5 Black (B). Total 10 balls.

Define $P(R_1)$ and $P(B_1)$:$P(R_1) = 5/10 = 1/2$$P(B_1) = 5/10 = 1/2$
Find conditional probabilities $P(R_2|R_1)$ and $P(R_2|B_1)$:
- If $R_1$ occurs, 2 R balls are added. Urn: 7 R, 5 B. Total 12.$P(R_2|R_1) = 7/12$
- If $B_1$ occurs, 2 B balls are added. Urn: 5 R, 7 B. Total 12.$P(R_2|B_1) = 5/12$
Apply Total Probability Theorem to find $P(R_2)$:$$P(R_2) = P(R_2|R_1)P(R_1) + P(R_2|B_1)P(B_1)$$$$P(R_2) = \left(\frac{7}{12}\right)\left(\frac{1}{2}\right) + \left(\frac{5}{12}\right)\left(\frac{1}{2}\right) = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \mathbf{1/2}$$

2. Ball is Red, Find Probability it’s from Bag I (Bayes’ Theorem)

Bag	Red (R)	Black (B)	Total
Bag I ($B_1$)	4	4	8
Bag II ($B_2$)	2	6	8

Let $B_1$ be the event that Bag I is selected, $B_2$ be the event that Bag II is selected.

Let $R$ be the event that the drawn ball is Red. We need $P(B_1|R)$.

Define $P(B_1)$ and $P(B_2)$:$P(B_1) = 1/2$$P(B_2) = 1/2$
Find $P(R|B_1)$ and $P(R|B_2)$:$P(R|B_1) = 4/8 = 1/2$$P(R|B_2) = 2/8 = 1/4$
Apply Bayes’ Theorem:

$$P(B_1|R) = \frac{P(R|B_1)P(B_1)}{P(R|B_1)P(B_1) + P(R|B_2)P(B_2)}$$
$$P(B_1|R) = \frac{(1/2)(1/2)}{(1/2)(1/2) + (1/4)(1/2)} = \frac{1/4}{1/4 + 1/8} = \frac{1/4}{3/8}$$
$$P(B_1|R) = \frac{1}{4} \times \frac{8}{3} = \mathbf{2/3}$$

3. Student has A Grade, Find Probability they are a Hostlier (Bayes’ Theorem)

Let $H$ be the event that the student is a Hostlier, $D$ be the event that the student is a Day Scholar.

Let $A$ be the event that the student attains an A grade. We need $P(H|A)$.

Define $P(H)$ and $P(D)$:$P(H) = 60\% = 0.60$$P(D) = 40\% = 0.40$
Find $P(A|H)$ and $P(A|D)$:$P(A|H) = 30\% = 0.30$$P(A|D) = 20\% = 0.20$
Apply Bayes’ Theorem:$$P(H|A) = \frac{P(A|H)P(H)}{P(A|H)P(H) + P(A|D)P(D)}$$$$P(H|A) = \frac{(0.30)(0.60)}{(0.30)(0.60) + (0.20)(0.40)} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26}$$$$P(H|A) = 18/26 = \mathbf{9/13}$$

4. Student Answers Correctly, Find Probability they Knew the Answer (Bayes’ Theorem)

Let $K$ be the event the student knows the answer, $G$ be the event the student guesses the answer.

Let $C$ be the event the answer is Correct. We need $P(K|C)$.

Define $P(K)$ and $P(G)$:$P(K) = 3/4$$P(G) = 1/4$
Find $P(C|K)$ and $P(C|G)$:
- If the student knows the answer, the probability of being correct is 1.$P(C|K) = 1$
- If the student guesses, the probability of being correct is $1/4$ (for a 4-option MC question).$P(C|G) = 1/4$
Apply Bayes’ Theorem:$$P(K|C) = \frac{P(C|K)P(K)}{P(C|K)P(K) + P(C|G)P(G)}$$$$P(K|C) = \frac{(1)(3/4)}{(1)(3/4) + (1/4)(1/4)} = \frac{3/4}{3/4 + 1/16} = \frac{12/16}{12/16 + 1/16}$$$$P(K|C) = \frac{12/16}{13/16} = \mathbf{12/13}$$

5. Test is Positive, Find Probability the Person Has the Disease (Bayes’ Theorem)

Let $D$ be the event the person has the disease, $H$ be the event the person is healthy ($H = D’$).

Let $P$ be the event the test is Positive. We need $P(D|P)$.

Define $P(D)$ and $P(H)$ (Prior Probabilities):$P(D) = 0.1\% = 0.001$$P(H) = 1 – 0.001 = 0.999$
Find $P(P|D)$ and $P(P|H)$ (Conditional Probabilities/Test Effectiveness):
- $P(P|D)$: Test is positive given disease is present (True Positive Rate).$P(P|D) = 99\% = 0.99$
- $P(P|H)$: Test is positive given person is healthy (False Positive Rate).$P(P|H) = 0.5\% = 0.005$
Apply Bayes’ Theorem:$$P(D|P) = \frac{P(P|D)P(D)}{P(P|D)P(D) + P(P|H)P(H)}$$$$P(D|P) = \frac{(0.99)(0.001)}{(0.99)(0.001) + (0.005)(0.999)}$$$$P(D|P) = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985}$$$$P(D|P) = \frac{990}{5985} = \frac{198}{1197} \approx \mathbf{0.165}$$

6. Coin Shows Heads, Find Probability it was the Two-Headed Coin (Bayes’ Theorem)

Let $C_1$ be the Two-Headed Coin, $C_2$ be the Biased Coin ($P(H)=0.75$), $C_3$ be the Unbiased Coin ($P(H)=0.5$).

Let $H$ be the event that the toss shows Heads. We need $P(C_1|H)$.

Define $P(C_1)$, $P(C_2)$, and $P(C_3)$:$P(C_1) = 1/3$$P(C_2) = 1/3$$P(C_3) = 1/3$
Find $P(H|C_1)$, $P(H|C_2)$, and $P(H|C_3)$:$P(H|C_1) = 1$$P(H|C_2) = 0.75 = 3/4$$P(H|C_3) = 0.5 = 1/2$
Apply Bayes’ Theorem:$$P(C_1|H) = \frac{P(H|C_1)P(C_1)}{P(H|C_1)P(C_1) + P(H|C_2)P(C_2) + P(H|C_3)P(C_3)}$$$$P(C_1|H) = \frac{(1)(1/3)}{(1)(1/3) + (3/4)(1/3) + (1/2)(1/3)}$$We can cancel $1/3$ from numerator and denominator:$$P(C_1|H) = \frac{1}{1 + 3/4 + 1/2} = \frac{1}{4/4 + 3/4 + 2/4} = \frac{1}{9/4}$$$$P(C_1|H) = \mathbf{4/9}$$

7. Person Meets Accident, Find Probability they are a Scooter Driver (Bayes’ Theorem)

Total insured persons $N = 2000 + 4000 + 6000 = 12000$.

Let $S$, $C$, $T$ be the events of a Scooter, Car, or Truck driver, respectively.

Let $A$ be the event of an Accident. We need $P(S|A)$.

Define $P(S)$, $P(C)$, and $P(T)$:$P(S) = 2000/12000 = 2/12 = 1/6$$P(C) = 4000/12000 = 4/12 = 2/6 = 1/3$$P(T) = 6000/12000 = 6/12 = 3/6 = 1/2$
Find $P(A|S)$, $P(A|C)$, and $P(A|T)$:$P(A|S) = 0.01 = 1/100$$P(A|C) = 0.03 = 3/100$$P(A|T) = 0.15 = 15/100 = 3/20$
Apply Bayes’ Theorem:$$P(S|A) = \frac{P(A|S)P(S)}{P(A|S)P(S) + P(A|C)P(C) + P(A|T)P(T)}$$$$P(S|A) = \frac{(1/100)(1/6)}{(1/100)(1/6) + (3/100)(1/3) + (15/100)(1/2)}$$Multiply numerator and denominator by 600 to clear fractions:$$P(S|A) = \frac{10}{10 + 60 + 450} = \frac{10}{520} = \mathbf{1/52}$$

8. Item is Defective, Find Probability it was Produced by Machine B (Bayes’ Theorem)

Let $A$ and $B$ be the events that the item was produced by Machine A or B.

Let $D$ be the event that the item is Defective. We need $P(B|D)$.

Define $P(A)$ and $P(B)$:$P(A) = 60\% = 0.60$$P(B) = 40\% = 0.40$
Find $P(D|A)$ and $P(D|B)$:$P(D|A) = 2\% = 0.02$$P(D|B) = 1\% = 0.01$
Apply Bayes’ Theorem:$$P(B|D) = \frac{P(D|B)P(B)}{P(D|A)P(A) + P(D|B)P(B)}$$$$P(B|D) = \frac{(0.01)(0.40)}{(0.02)(0.60) + (0.01)(0.40)}$$$$P(B|D) = \frac{0.0040}{0.0120 + 0.0040} = \frac{0.0040}{0.0160} = \frac{4}{16} = \mathbf{1/4}$$

9. New Product Introduced, Find Probability it was by the Second Group (Bayes’ Theorem)

Let $G_1$ and $G_2$ be the events that the First or Second group wins.

Let $N$ be the event that a New Product is Introduced. We need $P(G_2|N)$.

Define $P(G_1)$ and $P(G_2)$:$P(G_1) = 0.6$$P(G_2) = 0.4$
Find $P(N|G_1)$ and $P(N|G_2)$:$P(N|G_1) = 0.7$$P(N|G_2) = 0.3$
Apply Bayes’ Theorem:$$P(G_2|N) = \frac{P(N|G_2)P(G_2)}{P(N|G_1)P(G_1) + P(N|G_2)P(G_2)}$$$$P(G_2|N) = \frac{(0.3)(0.4)}{(0.7)(0.6) + (0.3)(0.4)}$$$$P(G_2|N) = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54}$$$$P(G_2|N) = 12/54 = \mathbf{2/9}$$

10. Girl Obtains Exactly One Head, Find Probability she Threw 1, 2, 3 or 4 (Bayes’ Theorem)

Let $E_1$ be the event the die shows 5 or 6. $P(E_1) = 2/6 = 1/3$. (Coin tossed 3 times)

Let $E_2$ be the event the die shows 1, 2, 3, or 4. $P(E_2) = 4/6 = 2/3$. (Coin tossed 1 time)

Let $H_1$ be the event of exactly one head. We need $P(E_2|H_1)$.

Find $P(H_1|E_1)$: Coin tossed 3 times ($S=\{HHH, \dots, TTT\}$).$P(\text{Heads}) = 1/2$.$P(H_1|E_1) = P(\text{Exactly 1 Head in 3 tosses}) = \binom{3}{1} (1/2)^1 (1/2)^2 = 3/8$
Find $P(H_1|E_2)$: Coin tossed 1 time ($S=\{H, T\}$).$P(H_1|E_2) = P(\text{Exactly 1 Head in 1 toss}) = P(H) = 1/2$
Apply Bayes’ Theorem:$$P(E_2|H_1) = \frac{P(H_1|E_2)P(E_2)}{P(H_1|E_1)P(E_1) + P(H_1|E_2)P(E_2)}$$$$P(E_2|H_1) = \frac{(1/2)(2/3)}{(3/8)(1/3) + (1/2)(2/3)} = \frac{1/3}{3/24 + 1/3} = \frac{1/3}{1/8 + 8/24}$$$$P(E_2|H_1) = \frac{1/3}{11/24} = \frac{1}{3} \times \frac{24}{11} = \mathbf{8/11}$$

11. Defective Item, Find Probability it was Produced by A (Bayes’ Theorem)

Let $A, B, C$ be the events the item was produced by operator A, B, or C.

Let $D$ be the event the item is Defective. We need $P(A|D)$.

Define $P(A)$, $P(B)$, and $P(C)$:$P(A) = 0.50$$P(B) = 0.30$$P(C) = 0.20$
Find $P(D|A)$, $P(D|B)$, and $P(D|C)$:$P(D|A) = 0.01$$P(D|B) = 0.05$$P(D|C) = 0.07$
Apply Bayes’ Theorem:$$P(A|D) = \frac{P(D|A)P(A)}{P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)}$$$$P(A|D) = \frac{(0.01)(0.50)}{(0.01)(0.50) + (0.05)(0.30) + (0.07)(0.20)}$$$$P(A|D) = \frac{0.0050}{0.0050 + 0.0150 + 0.0140} = \frac{0.0050}{0.0340}$$$$P(A|D) = 5/34$$

12. Two Drawn Cards are Diamonds, Find Probability the Lost Card was a Diamond (Bayes’ Theorem)

Total cards = 52. Diamonds (D) = 13. Non-Diamonds (ND) = 39.

Let $L_D$ be the event the lost card was a Diamond, $L_{ND}$ be the event the lost card was a Non-Diamond.

Let $A$ be the event that the two drawn cards (from the remaining 51) are both Diamonds. We need $P(L_D|A)$.

Define $P(L_D)$ and $P(L_{ND})$:$P(L_D) = 13/52 = 1/4$$P(L_{ND}) = 39/52 = 3/4$
Find $P(A|L_D)$ and $P(A|L_{ND})$:
- If $L_D$ occurs, remaining cards: 12 D, 39 ND. Total 51.$P(A|L_D) = P(\text{2 D from 51}) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{(12 \times 11)/(2 \times 1)}{(51 \times 50)/(2 \times 1)} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}$
- If $L_{ND}$ occurs, remaining cards: 13 D, 38 ND. Total 51.$P(A|L_{ND}) = P(\text{2 D from 51}) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}$
Apply Bayes’ Theorem:$$P(L_D|A) = \frac{P(A|L_D)P(L_D)}{P(A|L_D)P(L_D) + P(A|L_{ND})P(L_{ND})}$$$$P(L_D|A) = \frac{(132/2550)(1/4)}{(132/2550)(1/4) + (156/2550)(3/4)}$$Multiply numerator and denominator by $(2550 \times 4)$:$$P(L_D|A) = \frac{132}{132 + (156 \times 3)} = \frac{132}{132 + 468} = \frac{132}{600}$$$$P(L_D|A) = \frac{11}{50}$$

13. A Reports Head, Find Probability that Actually Head Appeared (Bayes’ Theorem)

Let $H$ be the event that a Head appears, $T$ be the event that a Tail appears.

Let $A_H$ be the event that A reports a Head.

$P(H) = 1/2$, $P(T) = 1/2$. $P(A \text{ speaks truth}) = 4/5$.

$P(A \text{ lies}) = 1 – 4/5 = 1/5$. We need $P(H|A_H)$.

Find $P(A_H|H)$ and $P(A_H|T)$:
- $P(A_H|H)$: Actual outcome is Head AND A speaks truth. $P(A_H|H) = 4/5$
- $P(A_H|T)$: Actual outcome is Tail AND A lies (reports Head). $P(A_H|T) = 1/5$
Apply Bayes’ Theorem:$$P(H|A_H) = \frac{P(A_H|H)P(H)}{P(A_H|H)P(H) + P(A_H|T)P(T)}$$$$P(H|A_H) = \frac{(4/5)(1/2)}{(4/5)(1/2) + (1/5)(1/2)} = \frac{4/10}{4/10 + 1/10} = \frac{4/10}{5/10} = 4/5$$

Answer: (A) $4/5$

14. If $A \subset B$ and $P(B) \ne 0$, then which of the following is correct?

Since $A \subset B$, the event $A$ is a subset of event $B$. This means that whenever $A$ occurs, $B$ must also occur.

Therefore, the intersection $A \cap B$ is simply $A$.

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$$

Since $A \subset B$ and $P(B) \ne 0$, we have $P(A) \le P(B)$, so $P(A)/P(B) \le 1$.

(A) $P(B) P(A|B)/P(A) = P(B) \cdot (P(A)/P(B)) / P(A) = P(A)/P(A) = 1$. This equation is mathematically true, but it is not one of the provided forms (it is not checking for equality with 1).
Consider the relationship between $P(A|B)$ and $P(A)$. Since $P(B) \le 1$ and $P(B) \ne 0$, we have $1/P(B) \ge 1$.$$P(A|B) = P(A) \cdot \frac{1}{P(B)}$$Since $1/P(B) \ge 1$, we must have $P(A|B) \ge P(A)$.

Answer: (C) $P(A|B) \ge P(A)$