A) \[\frac{3}{20}\]
B) \[\frac{21}{40}\]
C) \[\frac{3}{8}\]
D) None of these
Correct Answer: B
Solution :
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 \[A)]\] \[={{p}_{1}}\times {{p}_{4}}+{{p}_{2}}\times {{p}_{3}}=\frac{3}{8}\times \frac{4}{10}+\frac{5}{8}\times \frac{6}{10}=\frac{21}{40}\].
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