A) \[\frac{2}{7}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{7}\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
It is based on Baye?s theorem. Probability of picked bag A \[P(A)=\frac{1}{2}\] Probability of picked bag B \[P(B)=\frac{1}{2}\] Probability of green ball picked from bag A \[=P(A).P\left( \frac{G}{A} \right)\]\[=\frac{1}{2}\times \frac{4}{7}=\frac{2}{7}\] Probability of green ball picked from bag B \[=P(B).P\left( \frac{G}{B} \right)\]\[=\frac{1}{2}\times \frac{3}{7}=\frac{3}{14}\] Total probability of green ball = \[\frac{2}{7}+\frac{3}{14}=\frac{1}{2}\] Probability of fact that green ball is drawn from bag B \[P\left( \frac{G}{B} \right)=\frac{P(B)P\left( \frac{G}{B} \right)}{P(A)P\left( \frac{G}{A} \right)+P(B)P\left( \frac{G}{B} \right)}=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{3}{7}}=\frac{3}{7}\].You need to login to perform this action.
You will be redirected in
3 sec