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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 be white is
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$\frac{8}{15}$
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, then $P(A)=1 / 2, P(B)=1 / 2 \quad P(E / A)=2 / 5 \quad P(E / B)=4 / 6=2 / 3$
$P(E)=P(A) P(E / A)+P(B) P(E / B)=\frac{1}{2} \cdot \frac{2}{5}+\frac{1}{2} \cdot \frac{2}{3}=\frac{8}{15}$
$P(E)=P(A) P(E / A)+P(B) P(E / B)=\frac{1}{2} \cdot \frac{2}{5}+\frac{1}{2} \cdot \frac{2}{3}=\frac{8}{15}$
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