A purse contains 4 copper coins and 3 silver | vedclass

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Question

(C) Let $P(A)$ be the probability of selecting the first purse and $P(B)$ be the probability of selecting the second purse. Since there are two purses

Vedclass · Accepted Answer

$\frac{37}{56}$

Vedclass · Answer

(C) Let $P(A)$ be the probability of selecting the first purse and $P(B)$ be the probability of selecting the second purse. Since there are two purses,$P(A) = P(B) = \frac{1}{2}$.Let $C$ be the event of drawing a copper coin.The probability of drawing a copper coin from the first purse is $P(C|A) = \frac{4}{4+3} = \frac{4}{7}$.The probability of drawing a copper coin from the second purse is $P(C|B) = \frac{6}{6+2} = \frac{6}{8} = \frac{3}{4}$.Using the law of total probability,$P(C) = P(A) 	imes P(C|A) + P(B) 	imes P(C|B)$.$P(C) = \frac{1}{2} 	imes \frac{4}{7} + \frac{1}{2} 	imes \frac{3}{4} = \frac{2}{7} + \frac{3}{8}$.$P(C) = \frac{16+21}{56} = \frac{37}{56}$.

$A$ purse contains $4$ copper coins and $3$ silver coins,the second purse contains $6$ copper coins and $2$ silver coins. If a coin is drawn out of one of these purses,then the probability that it is a copper coin is :-

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