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  • A bag contains 5 red and 3 black balls and another bag contains 2 red and 6 black balls. Two balls are drawn at random (without replacement) from one of the bags and both are found to be red. Find the probability that balls are drawn from the first bag

A bag contains 5 red and 3 black balls and another bag contains 2 red and 6 black balls. Two balls are drawn at random (without replacement) from one of the bags and both are found to be red. Find the probability that balls are drawn from the first bag.

 

 

 

 
 
 
 
 

Answers (1)

Let E1: the ball is drawn from the first bag,
      E2: the ball is drawn from the second bag
      A: both the balls drawn are red
Here P\left ( E_{1} \right )= P\left ( E_{2} \right )= \frac{1}{2}\; \; P\left ( \frac{A}{E_{1}} \right )= \frac{5}{8}\times \frac{4}{7}= \frac{20}{56}
       P(\frac{A}{E_{2}})= \frac{2}{8}\times \frac{1}{7}= \frac{2}{56}
By Baye's theorem P\left ( \frac{E_{1}}{A} \right )= \frac{P\left ( E_{1} \right )P\left ( \frac{A}{E_{1}} \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 )}
\Rightarrow P\left ( \frac{E_{1}}{A} \right )= \frac{\frac{1}{2}\times \frac{20}{56}}{\frac{1}{2}\times \frac{20}{56}+\frac{1}{2}\times \frac{2}{56}}= \frac{10}{11}  

Posted by

Ravindra Pindel

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