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Q. 16  Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

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Let E1 and E2 respectively denote the event that red ball is transferred from bag 1 to bag 2 and a black ball is transferred from bag 1 to bag2.

  P(E1)=\frac{3}{7}          and      P(E2)=\frac{4}{7}

Let A be the event that ball drawn is red.

When a red ball is transferred from bag 1 to bag 2.

P(A|E1)=\frac{5}{10}=\frac{1}{2}

When a black ball is transferred from bag 1 to bag 2.

P(A|E2)=\frac{4}{10}=\frac{2}{5}

               P(E2|A)=\frac{P(E2).P(A|E2)}{P(E2).P(A|E2)+P(E1).P(A|E1)}

                                   =\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{4}{7}\times \frac{2}{5}+\frac{3}{7}\times \frac{1}{2}}

                                  =\frac{16}{31}

Posted by

seema garhwal

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