Get comprehensive solutions for the NCERT Class 12 Maths Chapter 13 Miscellaneous Exercise. Practice advanced problems covering Bayes’ Theorem

This exercise reviews key concepts from Probability, including Conditional Probability, Independence, Bayes’ Theorem, and Binomial Distribution.

Ex. 13.1

Ex. 13.2

Ex. 13.3

Ch 13 Misc.

1. Conditional Probability $P(B|A)$ where $P(A) \ne 0$

The formula for conditional probability is $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

(i) $A$ is a subset of $B$ ($A \subset B$)

If $A$ is a subset of $B$, then whenever $A$ occurs, $B$ must also occur. Therefore, the intersection $A \cap B$ is equal to $A$.

$$A \cap B = A \implies P(A \cap B) = P(A)$$

$$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = \mathbf{1}$$

(ii) $A \cap B = \phi$ (Mutually Exclusive)

If $A$ and $B$ are mutually exclusive, their intersection is the null set.

$$P(A \cap B) = 0$$

$$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)} = \mathbf{0}$$

2. Probability with Two Children

The sample space for two children is $S = \{MM, MF, FM, FF\}$, with $|S|=4$.

(i) Both children are males ($E$), if at least one is male ($F$)

$E$: Both are males $\implies E = \{MM\}$. $P(E) = 1/4$.
$F$: At least one is male $\implies F = \{MM, MF, FM\}$. $P(F) = 3/4$.
$E \cap F = \{MM\}$. $P(E \cap F) = 1/4$.$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{3/4} = \mathbf{1/3}$$

(ii) Both children are females ($E$), if the elder child is a female ($F$)

$E$: Both are females $\implies E = \{FF\}$. $P(E) = 1/4$.
$F$: Elder child is female $\implies F = \{FM, FF\}$. $P(F) = 2/4 = 1/2$.
$E \cap F = \{FF\}$. $P(E \cap F) = 1/4$.$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/4}{1/2} = \mathbf{1/2}$$

3. Grey Haired Person is Male (Bayes’ Theorem)

Let $M$ be the event the person is Male, $W$ be the event the person is Female.

Let $G$ be the event the person has Grey Hair. We need $P(M|G)$.

Prior Probabilities: Assume equal number of males and females.$$P(M) = 1/2, \quad P(W) = 1/2$$
Conditional Probabilities (Likelihood):$$P(G|M) = 5\% = 0.05$$$$P(G|W) = 0.25\% = 0.0025$$
Apply Bayes’ Theorem:$$P(M|G) = \frac{P(G|M)P(M)}{P(G|M)P(M) + P(G|W)P(W)}$$$$P(M|G) = \frac{(0.05)(1/2)}{(0.05)(1/2) + (0.0025)(1/2)}$$Cancel $1/2$ from numerator and denominator:$$P(M|G) = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525} = \frac{500}{525}$$$$P(M|G) = \mathbf{20/21}$$

4. At Most 6 of 10 People are Right-Handed (Binomial Distribution)

This is a binomial distribution problem: $n=10$ trials (people), $p=0.9$ (probability of being right-handed, $R$).

Let $X$ be the number of right-handed people. We need $P(X \le 6)$.

$$P(X \le 6) = 1 – P(X > 6) = 1 – [P(X=7) + P(X=8) + P(X=9) + P(X=10)]$$

The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$, where $q = 1 – p = 0.1$.

$P(X=7) = \binom{10}{7} (0.9)^7 (0.1)^3 = 120 \times 0.4782969 \times 0.001 \approx 0.0573956$
$P(X=8) = \binom{10}{8} (0.9)^8 (0.1)^2 = 45 \times 0.43046721 \times 0.01 \approx 0.1937092$
$P(X=9) = \binom{10}{9} (0.9)^9 (0.1)^1 = 10 \times 0.387420489 \times 0.1 \approx 0.3874205$
$P(X=10) = \binom{10}{10} (0.9)^{10} (0.1)^0 = 1 \times 0.3486784401 \times 1 \approx 0.3486784$

$$P(X > 6) \approx 0.0573956 + 0.1937092 + 0.3874205 + 0.3486784 = 0.9872037$$

$$P(X \le 6) = 1 – P(X > 6) \approx 1 – 0.9872037 = \mathbf{0.0127963}$$

5. Leap Year has 53 Tuesdays

A leap year has $366$ days, which is $52$ full weeks and $2$ extra days: $366 = 52 \times 7 + 2$.

The two extra days can be any of the following 7 equally likely pairs:

$S = \{(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)\}$

Let $E$ be the event that the leap year contains 53 Tuesdays. This occurs if the pair of extra days includes a Tuesday.

$$E = \{(Mon, Tue), (Tue, Wed)\}$$

$$P(E) = |E| / |S| = \mathbf{2/7}$$

6. Marble is Red, Find Probability it was from Box $A, B, C$ (Bayes’ Theorem)

| Box | Red (R) | White (W) | Black (B) | Total | $P(R|\text{Box}_i)$ |

| :— | :— | :— | :— | :— | :— |

| A | 1 | 6 | 3 | 10 | $1/10$ |

| B | 6 | 2 | 2 | 10 | $6/10$ |

| C | 8 | 1 | 1 | 10 | $8/10$ |

| D | 0 | 6 | 4 | 10 | $0/10$ |

Let $A, B, C, D$ be the events that the corresponding box is selected.

Since one box is selected at random, $P(A) = P(B) = P(C) = P(D) = 1/4$.

Let $R$ be the event that the drawn marble is Red.

$$P(R) = P(R|A)P(A) + P(R|B)P(B) + P(R|C)P(C) + P(R|D)P(D)$$

$$P(R) = (1/10)(1/4) + (6/10)(1/4) + (8/10)(1/4) + (0/10)(1/4)$$

$$P(R) = \frac{1}{40} + \frac{6}{40} + \frac{8}{40} + 0 = \frac{15}{40} = 3/8$$

We need $P(A|R)$, $P(B|R)$, $P(C|R)$.

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

7. Patient has Heart Attack, Find Probability they Followed Meditation (Bayes’ Theorem)

Let $M$ be the event of choosing Meditation/Yoga, $D$ be the event of choosing Drug Prescription.

Let $H$ be the event of suffering a Heart Attack. We need $P(M|H)$.

Prior Probabilities: Patient chooses any option with equal probabilities.$$P(M) = 1/2, \quad P(D) = 1/2$$
Conditional Probabilities $P(H|M)$ and $P(H|D)$:
- Initial risk of heart attack $= 40\% = 0.40$.
- $P(H|M)$: Meditation reduces risk by 30%. New risk is $0.40 \times (1 – 0.30) = 0.40 \times 0.70 = 0.28$.
- $P(H|D)$: Drug reduces risk by 25%. New risk is $0.40 \times (1 – 0.25) = 0.40 \times 0.75 = 0.30$.
Apply Bayes’ Theorem:$$P(M|H) = \frac{P(H|M)P(M)}{P(H|M)P(M) + P(H|D)P(D)}$$$$P(M|H) = \frac{(0.28)(1/2)}{(0.28)(1/2) + (0.30)(1/2)}$$Cancel $1/2$ from numerator and denominator:$$P(M|H) = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58} = 28/58$$$$P(M|H) = \mathbf{14/29}$$

8. Probability that a Second Order Determinant is Positive

A second order determinant is given by $A = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc$.

The elements $a, b, c, d \in \{0, 1\}$. There are $2^4 = 16$ possible determinants, each with probability $(1/2)^4 = 1/16$.

We require $ad – bc > 0$, or $ad > bc$.

Since $a, b, c, d \in \{0, 1\}$, the possible values for $ad$ and $bc$ are $\{0, 1\}$.

The condition $ad > bc$ is satisfied only when:

$ad = 1$ and $bc = 0$

Case 1: $ad = 1$This requires $a=1$ AND $d=1$. (1 way)
Case 2: $bc = 0$This requires $b=0$ OR $c=0$ (or both).Total possibilities for $(b, c)$ are $(0, 0), (0, 1), (1, 0), (1, 1)$.$bc=0$ for all but $(1, 1)$. (3 ways: $(0, 0), (0, 1), (1, 0)$)

For $ad > bc$, we must have $a=1, d=1$ AND $bc=0$.

Number of favorable outcomes $= (\text{ways for } ad=1) \times (\text{ways for } bc=0)$

Number of favorable outcomes $= 1 \times 3 = 3$.

The favorable outcomes are: $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$, $\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}$, $\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}$.

The value of the determinants are $1, 1, 1$, respectively, which are all positive.

Probability that the value is positive $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \mathbf{3/16}$

9. Electronic Assembly Failures (Set Theory Probability)

Let $A$ be the event ‘A fails’, $B$ be the event ‘B fails’.

Given:

$P(A) = 0.2$
$P(B \text{ fails alone}) = P(B \cap A’) = 0.15$
$P(A \text{ and } B \text{ fail}) = P(A \cap B) = 0.15$

(i) $P(A \text{ fails}|B \text{ has failed}) = P(A|B)$

To find $P(A|B)$, we first need $P(B)$.

We know that $P(B) = P(B \cap A’) + P(B \cap A)$ (Since $A’$ and $A$ are mutually exclusive).

$$P(B) = P(B \text{ fails alone}) + P(A \text{ and } B \text{ fail})$$

$$P(B) = 0.15 + 0.15 = 0.30$$

Now apply conditional probability:

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.15}{0.30} = \mathbf{1/2} \text{ or } \mathbf{0.5}$$

(ii) $P(A \text{ fails alone}) = P(A \cap B’)$

$P(A \text{ fails alone}) = P(A) – P(A \cap B)$

$$P(A \cap B’) = 0.2 – 0.15 = \mathbf{0.05}$$

10. Ball Drawn is Red, Find Probability the Transferred Ball was Black (Bayes’ Theorem)

Initial: Bag I (3R, 4B, Total 7). Bag II (4R, 5B, Total 9).

One ball is transferred from I to II. Let $T_R$ be event Red transferred, $T_B$ be event Black transferred.

Let $R_{II}$ be the event Red is drawn from Bag II. We need $P(T_B|R_{II})$.

Prior Probabilities:$$P(T_R) = 3/7$$$$P(T_B) = 4/7$$
Conditional Probabilities:
- If $T_R$: Bag II has (4+1=5) R, 5 B. Total 10. $P(R_{II}|T_R) = 5/10 = 1/2$.
- If $T_B$: Bag II has 4 R, (5+1=6) B. Total 10. $P(R_{II}|T_B) = 4/10 = 2/5$.
Apply Bayes’ Theorem:$$P(T_B|R_{II}) = \frac{P(R_{II}|T_B)P(T_B)}{P(R_{II}|T_R)P(T_R) + P(R_{II}|T_B)P(T_B)}$$$$P(T_B|R_{II}) = \frac{(2/5)(4/7)}{(1/2)(3/7) + (2/5)(4/7)}$$Multiply numerator and denominator by 35 to clear fractions:$$P(T_B|R_{II}) = \frac{2 \times 4 \times 5}{1 \times 3 \times 5 + 2 \times 4 \times 7} \quad (\text{Error in cancelling } 7 \text{s and } 5\text{s})$$$$P(T_B|R_{II}) = \frac{8/35}{3/14 + 8/35}$$Find common denominator for 14 and 35 (which is 70):$$P(T_B|R_{II}) = \frac{8/35}{(15/70 + 16/70)} = \frac{8/35}{31/70}$$$$P(T_B|R_{II}) = \frac{8}{35} \times \frac{70}{31} = \frac{8 \times 2}{31} = \mathbf{16/31}$$

Multiple Choice Questions

11. If $P(A) \ne 0$ and $P(B|A) = 1$, then

$P(B|A) = \frac{P(A \cap B)}{P(A)} = 1$. This implies $P(A \cap B) = P(A)$.

Since the probability of the intersection is equal to the probability of $A$, it means that every outcome in $A$ is also in $B$.

This is the definition of $A$ being a subset of $B$.

$$\mathbf{A \subset B}$$

Answer: (A) $A \subset B$

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

The condition $P(A|B) > P(A)$ means the occurrence of $B$ makes $A$ more likely.

Since $P(A|B) = \frac{P(A \cap B)}{P(B)}$, we have $\frac{P(A \cap B)}{P(B)} > P(A)$, or $P(A \cap B) > P(A)P(B)$.

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

Since $P(A \cap B) > P(A)P(B)$, dividing by $P(A)$ (assuming $P(A) \ne 0$):

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

$$P(B|A) > P(B)$$

Answer: (C) $P(B|A) > P(B)$

13. If $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then

Given the equation $P(A) + P(B) – P(A \cap B) = P(A)$.

We know that $P(A) + P(B) – P(A \cap B) = P(A \cup B)$.

So, $P(A \cup B) = P(A)$.

Subtract $P(A)$ from both sides of the original equation:

$$P(B) – P(A \cap B) = 0 \implies P(A \cap B) = P(B)$$

Since $P(A \cap B) = P(B)$, every outcome in $B$ must also be in $A$, which means $B \subset A$.

Now, let’s evaluate the options using the result $P(A \cap B) = P(B)$:

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

Answer: (B) $P(A|B) = 1$