question_answer

Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B

Answer:

Sol.      Let \[{{E}_{1}}\] be the event of choosing bag A \[{{E}_{2}}\] be the event of choosing bag B and A be the event of drawing a red ball. Then,    \[P({{E}_{1}})=P({{E}_{2}})=\frac{1}{2}\] Also, \[P(A\,/{{E}_{1}})=\] Probability of choosing a red ball from bag \[A=\frac{3}{5}\] and \[P(A\,/{{E}_{2}})=\] Probability of choosing a red ball from bag \[B=\frac{5}{9}\] Now, the probability of drawing a ball from bag B, being given that it is red, is \[P({{E}_{2}}/A).\] By using Baye's theorem, we have \[P({{E}_{2}}/A)=\frac{P({{E}_{2}})\cdot P(A\,/{{E}_{2}})}{P({{E}_{1}})\cdot P(A\,/{{E}_{1}})+P({{E}_{2}})\cdot P(A\,/{{E}_{2}})}\]             \[=\frac{\frac{1}{2}\cdot \frac{5}{9}}{\frac{1}{2}\cdot \frac{3}{5}+\frac{1}{2}\cdot \frac{5}{9}}=\frac{\frac{5}{9}}{\frac{27+25}{45}}\]             \[=\frac{5}{9}\times \frac{45}{52}=\frac{25}{52}\]