A) \[\frac{1}{7}\]
B) \[\frac{2}{7}\]
C) \[\frac{3}{7}\]
D) \[\frac{4}{7}\]
Correct Answer: B
Solution :
Let\[P({{E}_{1}})=\]Probability that the ball is picked from the first bag\[=\frac{1}{2}\] \[P({{E}_{2}})=\]Probability that the ball is picked from the second bag\[=\frac{1}{2}\] Now, \[P(B/{{E}_{1}})=\frac{3}{7}\] (where P = probability that the picked ball is black) \[P(B/{{E}_{2}})=\frac{4}{7}\] Now, required probability\[P({{E}_{2}}/B)\] \[=\frac{P({{E}_{2}}).P(B/{{E}_{2}})}{P({{E}_{1}})P(B/{{E}_{1}})+P({{E}_{2}})P(B/{{E}_{2}})}\] \[=\frac{\frac{1}{2}.\frac{4}{7}}{\frac{1}{2}.\frac{3}{7}+\frac{1}{2}.\frac{4}{7}}=\frac{4}{7}\]
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