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  • A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red.

Q. 2   A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn frome bag which is found to be red. Find the probability that the ball is drawn from the first bag.

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BAG 1 : Red balls =4        Black balls=4     Total balls = 8                                                                                                          

BAG 2 : Red balls = 2       Black balls = 6     Total balls = 8                                                                              

B1 : selecting bag 1

B2 : selecting bag 2

P(B1)=P(B2)=\frac{1}{2}

Let R be a event of getting red ball

P(R|B1) = P(drawing\, \, red\, \, ball\, \, from \, first \, \, bag)= \frac{4}{8}=\frac{1}{2}

P(R|B2) = P(drawing\, \, red\, \, ball\, \, from \, second \, \, bag)= \frac{2}{8}=\frac{1}{4}

probability that the ball is drawn from the first bag,

given that it is red is P(B1|R).

Using Baye's theorem, we have

             P(B1|R) = \frac{P(B1).P(R|B1)}{P(B1).P(R|B1)+P(B2).P(R|B2)}

           P(B1|R) = \frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}

            P(B1|R) =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}

         P(B1|R) =\frac{\frac{1}{4}}{\frac{3}{8}}

        P(B1|R) = \frac{2}{3}

Posted by

seema garhwal

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