Q. 10    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 4with the die?

Answers (1)
S seema garhwal

Let, A: Outcome on die is 5 or 6.

       B: Outcome on die is 1,2,3,4

P(A)=\frac{2}{6}=\frac{1}{3}

P(B)=\frac{4}{6}=\frac{2}{3}

X: Event of getting exactly one head.

Probability of  getting exactly one head when she tosses a coin three times : P(X|A)=\frac{3}{8}

Probability of  getting exactly one head when she tosses a coin one time :  P(X|B)=\frac{1}{2}

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{2}{3}\times \frac{1}{2}}{\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{3}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{11}{24}}

P(B|X)= \frac{1\times 24}{3\times 11}=\frac{8}{11}

Hence, the probability that she threw 1,2,3 or 4 with the die = 

                                                                                                      P(B|X)=\frac{8}{11}

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