There are three coins. One is a two-headed co | vedclass

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Question

(A) Let $E_{1}$,$E_{2}$,and $E_{3}$ be the events of choosing a two-headed coin,a biased coin,and an unbiased coin,respectively.Since one coin is chos

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(A) Let $E_{1}$,$E_{2}$,and $E_{3}$ be the events of choosing a two-headed coin,a biased coin,and an unbiased coin,respectively.Since one coin is chosen at random,$P(E_{1}) = P(E_{2}) = P(E_{3}) = \frac{1}{3}$.Let $A$ be the event that the coin shows heads.For a two-headed coin,$P(A|E_{1}) = 1$.For a biased coin,$P(A|E_{2}) = 75\% = \frac{75}{100} = \frac{3}{4}$.For an unbiased coin,$P(A|E_{3}) = \frac{1}{2}$.Using Bayes' theorem,the probability that the coin is two-headed given that it shows heads is $P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2}) + P(E_{3})P(A|E_{3})}$.Substituting the values: $P(E_{1}|A) = \frac{\frac{1}{3} 	imes 1}{\frac{1}{3} 	imes 1 + \frac{1}{3} 	imes \frac{3}{4} + \frac{1}{3} 	imes \frac{1}{2}}$.$P(E_{1}|A) = \frac{\frac{1}{3}}{\frac{1}{3} (1 + \frac{3}{4} + \frac{1}{2})} = \frac{1}{\frac{4+3+2}{4}} = \frac{1}{\frac{9}{4}} = \frac{4}{9}$.

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