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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 $4$ with the die?

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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 =

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

Posted by

seema garhwal

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