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14.Probability
medium
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
A
$\frac{3}{{20}}$
B
$\frac{{21}}{{40}}$
C
$\frac{3}{8}$
D
None of these
Solution
(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} \times {p_4} + {p_2} \times {p_3} = \frac{3}{8} \times \frac{4}{{10}} + \frac{5}{8} \times \frac{6}{{10}} = \frac{{21}}{{40}}$.
Standard 11
Mathematics
