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Q10:
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Solution :

Step 1: Defining Events and Sample Spaces

Let $E_1$ be the event that the girl gets 5 or 6 on the die. The probability of this event is $P(E_1) = \frac{2}{6} = \frac{1}{3}$.

Let $E_2$ be the event that the girl gets 1, 2, 3, or 4 on the die. The probability of this event is $P(E_2) = \frac{4}{6} = \frac{2}{3}$.

Let $A$ be the event that she obtains exactly one head.

Step 2: Calculating Conditional Probabilities

If $E_1$ occurs (she gets 5 or 6), she tosses a coin 3 times. The number of heads follows a binomial distribution $B(n=3, p=0.5)$. The probability of getting exactly one head is:

$P(A|E_1) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \times \frac{1}{8} = \frac{3}{8}$

If $E_2$ occurs (she gets 1, 2, 3, or 4), she tosses a coin once. The probability of getting exactly one head is:

$P(A|E_2) = \binom{1}{1} \left(\frac{1}{2}\right)^1 = \frac{1}{2}$

Step 3: Visual Representation of Probability Tree

P(E1)=1/3 P(E2)=2/3 P(A|E1)=3/8 P(A|E2)=1/2 Start

Step 4: Applying Bayes' Theorem

We are required to find $P(E_2|A)$. According to Bayes' Theorem:

$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

Substituting the values calculated in steps 1 and 2:

$P(E_2|A) = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\left(\frac{1}{3} \cdot \frac{3}{8}\right) + \left(\frac{2}{3} \cdot \frac{1}{2}\right)}$

Step 5: Algebraic Simplification

Numerator: $\frac{2}{6} = \frac{1}{3}$

Denominator: $\left(\frac{3}{24}\right) + \left(\frac{2}{6}\right) = \frac{1}{8} + \frac{1}{3}$

Finding common denominator for the sum: $\frac{3 + 8}{24} = \frac{11}{24}$

$P(E_2|A) = \frac{1/3}{11/24} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}$

Final Solution: The probability that she threw 1, 2, 3, or 4 given that she obtained exactly one head is $\frac{8}{11}$.


More Questions from Class 12 Mathematics Probability EXERCISE 13.3


CBSE Solutions for Class 12 Mathematics Probability


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