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  • Bag I contains 4 red and 2 green balls and Bag II contains 3 red and 5 green balls. One ball is transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. The ball so drawn is found to be green in colour. Find the probab

Bag I contains 4 red and 2 green balls and Bag II contains 3 red and 5 green balls. One ball is transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. The ball so drawn is found to be green in colour. Find the probability that the transferred ball is also green.

 

 

 

 
 
 
 
 

Answers (1)

Let E1 and E2 respectively denote the events that a red ball is transferred from a bag I to II and a green ball is transferred from the bag I to II
p\left ( E_{1} \right )= \frac{4}{6}                          p\left ( E_{2} \right )= \frac{2}{6}
Let A be the event that the ball drawn is green
(1) when a red ball is transferred from the bag I and II
     p\left ( \frac{A}{E_{1}}\right )= \frac{5}{9}
(2) when a  green ball is transferred from the bag I and II
     p\left ( \frac{A}{E_{2}}\right )= \frac{6}{9}
\therefore p\left ( \frac{E_{2}}{A} \right )= \frac{p\left ( E_{2} \right )p\left ( \frac{A}{E_{2}} \right )}{p\left ( E_{1} \right )p\left ( \frac{A}{E_{1}} \right )+p\left ( E_{2} \right )p\left ( \frac{A}{E_{2}} \right )}
                    = \frac{\frac{2}{6}\times \frac{6}{9}}{\frac{4}{6}\times \frac{5}{9}+\frac{2}{6}\times \frac{6}{9}}\; \; = \frac{12}{32}=\frac{3}{8}
              

Posted by

Ravindra Pindel

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