Suppose a girl throws a die. If she gets a 5 | vedclass

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Question

(A) Let $E_{1}$ be the event that the outcome on the die is $5$ or $6$ and $E_{2}$ be the event that the outcome on the die is $1, 2, 3,$ or $4$.$P(E_

Vedclass · Accepted Answer

$\frac{8}{11}$

Vedclass · Answer

(A) Let $E_{1}$ be the event that the outcome on the die is $5$ or $6$ and $E_{2}$ be the event that the outcome on the die is $1, 2, 3,$ or $4$.$P(E_{1}) = \frac{2}{6} = \frac{1}{3}$ and $P(E_{2}) = \frac{4}{6} = \frac{2}{3}$.Let $A$ be the event of getting exactly one head.$P(A | E_{1})$ is the probability of getting exactly one head by tossing a coin three times: $\binom{3}{1} (\frac{1}{2})^1 (\frac{1}{2})^2 = \frac{3}{8}$.$P(A | E_{2})$ is the probability of getting exactly one head in a single throw of a coin: $\frac{1}{2}$.We need to find $P(E_{2} | A)$. By Bayes' theorem:$P(E_{2} | A) = \frac{P(E_{2}) P(A | E_{2})}{P(E_{1}) P(A | E_{1}) + P(E_{2}) P(A | E_{2})}$$P(E_{2} | A) = \frac{(\frac{2}{3}) (\frac{1}{2})}{(\frac{1}{3}) (\frac{3}{8}) + (\frac{2}{3}) (\frac{1}{2})}$$P(E_{2} | A) = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{3+8}{24}} = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{1}{3} 	imes \frac{24}{11} = \frac{8}{11}$.

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