A bag contains 16 coins,of which 2 are counte | vedclass

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(A) Let $C_1$ be the event of selecting a counterfeit coin and $C_2$ be the event of selecting a fair coin.Total coins = $16$.Number of counterfeit co

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(A) Let $C_1$ be the event of selecting a counterfeit coin and $C_2$ be the event of selecting a fair coin.Total coins = $16$.Number of counterfeit coins = $2$. Probability $P(C_1) = \frac{2}{16}$.Number of fair coins = $14$. Probability $P(C_2) = \frac{14}{16}$.For a counterfeit coin,the probability of getting a head $P(H|C_1) = 1$.For a fair coin,the probability of getting a head $P(H|C_2) = \frac{1}{2}$.Using the law of total probability,$P(H) = P(C_1) 	imes P(H|C_1) + P(C_2) 	imes P(H|C_2)$.$P(H) = \frac{2}{16} 	imes 1 + \frac{14}{16} 	imes \frac{1}{2} = \frac{2}{16} + \frac{7}{16} = \frac{9}{16}$.

$A$ bag contains $16$ coins,of which $2$ are counterfeit with heads on both sides. The rest are fair coins. One coin is selected at random from the bag and tossed. The probability of getting a head is

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