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(B) Let $E_1$ be the event of choosing bag $x$ and $E_2$ be the event of choosing bag $y$. Since the bag is chosen at random,$P(E_1) = P(E_2) = 1/2$.L

Vedclass · Accepted Answer

$7/15$

Vedclass · Answer

(B) Let $E_1$ be the event of choosing bag $x$ and $E_2$ be the event of choosing bag $y$. Since the bag is chosen at random,$P(E_1) = P(E_2) = 1/2$.Let $W$ be the event of picking a white ball.The probability of picking a white ball from bag $x$ is $P(W|E_1) = 3/5$.The probability of picking a white ball from bag $y$ is $P(W|E_2) = 2/6 = 1/3$.Using the law of total probability,$P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.$P(W) = (1/2) \cdot (3/5) + (1/2) \cdot (1/3) = 3/10 + 1/6 = (9 + 5) / 30 = 14/30 = 7/15$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

$A$ bag $x$ contains $3$ white balls and $2$ black balls and another bag $y$ contains $2$ white balls and $4$ black balls. $A$ bag and a ball out of it are picked at random. The probability that the ball is white,is

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