A purse contains 4 copper and 3 silver coins. | vedclass

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(D) Let $A$ be the event of choosing the first purse and $B$ be the event of choosing the second purse. Since there are two purses,$P(A) = 1/2$ and $P

Vedclass · Accepted Answer

$37/56$

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

$A$ purse contains $4$ copper and $3$ silver coins. $A$ second purse contains $6$ copper and $2$ silver coins. If a coin is drawn from any one of these two purses,what is the probability that it is a copper coin?

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