Two coins A and B are kept in an urn. When co | vedclass

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(D) Let $E$ be the event that the chosen coin shows heads on both tosses. Let $C_A$ and $C_B$ be the events of choosing coin $A$ and coin $B$ respecti

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$1/10$

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(D) Let $E$ be the event that the chosen coin shows heads on both tosses. Let $C_A$ and $C_B$ be the events of choosing coin $A$ and coin $B$ respectively.Given $P(C_A) = 1/2$ and $P(C_B) = 1/2$.The probability of getting heads twice given coin $A$ is $P(E|C_A) = (1/4)^2 = 1/16$.The probability of getting heads twice given coin $B$ is $P(E|C_B) = (3/4)^2 = 9/16$.By Bayes' theorem,the probability that the coin is $A$ given that it showed heads twice is:$P(C_A|E) = \frac{P(C_A)P(E|C_A)}{P(C_A)P(E|C_A) + P(C_B)P(E|C_B)}$$P(C_A|E) = \frac{(1/2)(1/16)}{(1/2)(1/16) + (1/2)(9/16)}$$P(C_A|E) = \frac{1/32}{1/32 + 9/32} = \frac{1}{1+9} = 1/10$.

Two coins $A$ and $B$ are kept in an urn. When coin $A$ is flipped,the probability of getting a Head is $1/4$,while for coin $B$ it is $3/4$. One coin is randomly chosen from this bag and tossed twice,and it falls heads on both occasions. The probability that it is coin $A$ is:

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