A bag P contains 5 white marbles and 3 black | vedclass

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(A) Let $W$ denote white marbles and $B$ denote black marbles. Bag $P$ contains $5W$ and $3B$. Four marbles are transferred to bag $Q$.Let $E_1$ be th

Vedclass · Accepted Answer

$\frac{1}{7}$

Vedclass · Answer

(A) Let $W$ denote white marbles and $B$ denote black marbles. Bag $P$ contains $5W$ and $3B$. Four marbles are transferred to bag $Q$.Let $E_1$ be the event that $1W$ and $3B$ are transferred.Let $E_2$ be the event that $2W$ and $2B$ are transferred.Let $E_3$ be the event that $3W$ and $1B$ are transferred.Let $E_4$ be the event that $4W$ and $0B$ are transferred.Let $A$ be the event that a black marble is drawn from bag $Q$.The total number of ways to choose $4$ marbles from $8$ is $^8C_4 = 70$.$P(E_1) = \frac{^5C_1 	imes ^3C_3}{70} = \frac{5 	imes 1}{70} = \frac{5}{70}$$P(E_2) = \frac{^5C_2 	imes ^3C_2}{70} = \frac{10 	imes 3}{70} = \frac{30}{70}$$P(E_3) = \frac{^5C_3 	imes ^3C_1}{70} = \frac{10 	imes 3}{70} = \frac{30}{70}$$P(E_4) = \frac{^5C_4 	imes ^3C_0}{70} = \frac{5 	imes 1}{70} = \frac{5}{70}$The conditional probabilities of drawing a black marble from $Q$ are:$P(A|E_1) = \frac{3}{4}, P(A|E_2) = \frac{2}{4}, P(A|E_3) = \frac{1}{4}, P(A|E_4) = 0$.Using Bayes' theorem,we find $P(E_1|A)$:$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{\sum_{i=1}^4 P(E_i)P(A|E_i)}$$P(E_1|A) = \frac{\frac{5}{70} 	imes \frac{3}{4}}{\frac{5}{70} 	imes \frac{3}{4} + \frac{30}{70} 	imes \frac{2}{4} + \frac{30}{70} 	imes \frac{1}{4} + \frac{5}{70} 	imes 0}$$P(E_1|A) = \frac{15}{15 + 60 + 30 + 0} = \frac{15}{105} = \frac{1}{7}$.

Box	Black	White	Red	Blue
$I$	$3$	$4$	$5$	$6$
$II$	$2$	$2$	$2$	$2$
$III$	$1$	$2$	$3$	$1$
$IV$	$4$	$3$	$1$	$5$

$A$ bag $P$ contains $5$ white marbles and $3$ black marbles. Four marbles are drawn at random from $P$ and are put in an empty bag $Q$. If a marble drawn at random from $Q$ is found to be black,then the probability that all the three black marbles in $P$ were transferred to the bag $Q$ is:

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