A bag contains 3 red and 5 black balls and a | vedclass

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Question

(b) From bag $A,\,\,P$ (red ball) $ = {p_1} = \frac{3}{8}$

$P($black ball)$ = {p_2} = \frac{5}{8}$

From bag $B,$ $P$ (red ball) $ = {p_3} = \fra

Vedclass · Accepted Answer

$\frac{{21}}{{40}}$

Vedclass · Answer

(b) From bag $A,\,\,P$ (red ball) $ = {p_1} = \frac{3}{8}$

$P($black ball)$ = {p_2} = \frac{5}{8}$

From bag $B,$ $P$ (red ball) $ = {p_3} = \frac{6}{{10}}$

$P$(black ball) = ${p_4} = \frac{4}{{10}}$

Required probability

$ = P$[(red ball from bag $A$ and black from $B)$ or

(red from bag $B$ and black from $A)]$

$ = {p_1} 	imes {p_4} + {p_2} 	imes {p_3} = \frac{3}{8} 	imes \frac{4}{{10}} + \frac{5}{8} 	imes \frac{6}{{10}} = \frac{{21}}{{40}}$.

A bag contains $3$ red and $5$ black balls and a second bag contains $6$ red and $4$ black balls. A ball is drawn from each bag. The probability that one is red and other is black, is

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