A purse contains 4 copper coins and 3 silver | vedclass

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Question

(A) Let $E_1$ be the event of choosing the first purse and $E_2$ be the event of choosing the second purse. Since the purse is chosen randomly,$P(E_1)

Vedclass · Accepted Answer

$\frac{41}{70}$

Vedclass · Answer

(A) Let $E_1$ be the event of choosing the first purse and $E_2$ be the event of choosing the second purse. Since the purse is chosen randomly,$P(E_1) = P(E_2) = \frac{1}{2}$.Let $A$ be the event of drawing a copper coin.For the first purse,$P(A|E_1) = \frac{4}{4+3} = \frac{4}{7}$.For the second purse,$P(A|E_2) = \frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$.Using the law of total probability,the probability of drawing a copper coin is:$P(A) = P(E_1) 	imes P(A|E_1) + P(E_2) 	imes P(A|E_2)$$P(A) = \frac{1}{2} 	imes \frac{4}{7} + \frac{1}{2} 	imes \frac{3}{5}$$P(A) = \frac{2}{7} + \frac{3}{10}$$P(A) = \frac{20 + 21}{70} = \frac{41}{70}$.

$A$ purse contains $4$ copper coins and $3$ silver coins. $A$ second purse contains $6$ copper coins and $4$ silver coins. $A$ purse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?

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