A) \[\frac{5}{14}\]
B) \[\frac{5}{16}\]
C) \[\frac{5}{18}\]
D) \[\frac{25}{52}\]
Correct Answer: D
Solution :
Let \[{{E}_{1}}\] be the event that the ball is drawn from bag \[A,\,{{E}_{2}}\] the event that it is drawn from bag \[B\] and \[E\] that the ball is red.We have to find \[P({{E}_{2}}/E)\]. Since both the bags are equally likely to be selected, we have \[P({{E}_{1}})=P({{E}_{2}})=\frac{1}{2}\] Also \[P(E/{{E}_{1}})=3/5\] and \[P(E/{{E}_{2}})=5/9.\] Hence by Bay?s theorem, we have \[P({{E}_{2}}/E)=\frac{P({{E}_{2}})\,P(E/{{E}_{2}})}{P({{E}_{1}})\,P(E/{{E}_{1}})+P({{E}_{2}})\,P(E/{{E}_{2}})}\] \[=\frac{\frac{1}{2}.\frac{5}{9}}{\frac{1}{2}.\frac{3}{5}+\frac{1}{2}.\frac{5}{9}}=\frac{25}{52}.\]
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