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  • A bag contains 5 red and 4 black balls,a second bag contains 3 red and 6 black balls.One of the two bags is selected at random and two balls are drawn at random(without replacement) both of which are found to be red. Find the probability that the balls

A bag contains 5 red and 4 black balls,a second bag contains 3 red and 6 black balls.One of the two bags is selected at random and two balls are drawn at random(without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag.

 

 

 

 
 
 
 
 

Answers (1)

Let E_{1}= event of choosing the bag 1
      E_{2}= be the event of choosing bag 2
A\Rightarrow be the event of drawing red ball
Then,
p\left ( E_{1} \right )= p\left ( E_{2} \right )= \frac{1}{2}
Now, p\left ( \frac{A}{E_{1}} \right )= \frac{5}{9}\times \frac{4}{8}= \frac{20}{72}
and  p\left ( \frac{A}{E_{2}} \right )= \frac{3}{9}\times \frac{2}{8}= \frac{6}{72}
Now,The probability of drawing a ball from bag 2
if it is given that it is red is p\left ( \frac{E_{2}}{A} \right ).
Now by Bayes thrown, we have
p\left ( \frac{E_{2}}{A} \right ).= \frac{p\left ( E_{2} \right )\cdot p\left ( \frac{A}{E_{2}} \right )}{p\left ( E_{1} \right )\cdot p\left ( \frac{A}{E_{1}} \right )+p\left ( E_{2} \right )p\left ( \frac{A}{E_{1}} \right )}
                  = \frac{\frac{1}{2}\times \frac{6}{72}}{\frac{1}{2}\times \frac{20}{72}+\frac{1}{2}\times \frac{6}{72}}\Rightarrow \frac{\frac{6}{2\times \not{72}}}{\frac{20}{2\times \not{72}}+\frac{6}{2\times \not{72}}}
                 \Rightarrow \frac{6}{20+6}\Rightarrow \frac{6}{26}\Rightarrow \frac{3}{13}

Posted by

Ravindra Pindel

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