A purse contains 4 copper coins and 3 silver | vedclass

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(C) Let $P(A)$ be the event of choosing the first purse and $P(B)$ be the event of choosing the second purse. Since the purse is chosen at random,$P(A

Vedclass · Accepted Answer

$37/56$

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(C) Let $P(A)$ be the event of choosing the first purse and $P(B)$ be the event of choosing the second purse. Since the purse is chosen at random,$P(A) = P(B) = 1/2$.Let $C$ be the event of drawing a copper coin.The probability of drawing a copper coin from the first purse is $P(C|A) = 4/(4+3) = 4/7$.The probability of drawing a copper coin from the second purse is $P(C|B) = 6/(6+2) = 6/8 = 3/4$.Using the Law of Total Probability,the total probability of drawing a copper coin is $P(C) = P(A) 	imes P(C|A) + P(B) 	imes P(C|B)$.$P(C) = (1/2 	imes 4/7) + (1/2 	imes 3/4) = 2/7 + 3/8$.To add these fractions,find the common denominator,which is $56$.$P(C) = (16/56) + (21/56) = 37/56$.

$A$ 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 any purse,then the probability that it is a copper coin is

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