A bag X contains 2 white and 3 black balls an | vedclass

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(C) Let $A$ be the event of selecting bag $X$,$B$ be the event of selecting bag $Y$,and $E$ be the event of drawing a white ball.Since one bag is sele

Vedclass · Accepted Answer

$\frac{8}{15}$

Vedclass · Answer

(C) Let $A$ be the event of selecting bag $X$,$B$ be the event of selecting bag $Y$,and $E$ be the event of drawing a white ball.Since one bag is selected at random,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{2}$.The probability of drawing a white ball from bag $X$ is $P(E|A) = \frac{2}{2+3} = \frac{2}{5}$.The probability of drawing a white ball from bag $Y$ is $P(E|B) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.Using the law of total probability,$P(E) = P(A)P(E|A) + P(B)P(E|B)$.$P(E) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

$A$ bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then the probability for the ball chosen to be white is

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