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Question
Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red find the probability that two red balls were transferred from A to B.
Solution
It is given that bag A contains 3 red and 5 black balls and bag B contains 4 red and 4 black balls.
Let E1, E2, E3 and A be the events as defined below:
E1 : Two red balls are transferred from bag A to bag B.
E2 : One red ball and one black ball is transferred from bag A to bag B.
E3 : Two black balls are transferred from bag A to bag B.
A : Ball drawn from bag B is red.
So,
\[P\left( E_1 \right) = \frac{^{3}{}{C}_2}{^{8}{}{C}_2} = \frac{3}{28}\]
\[P\left( E_2 \right) = \frac{^{3}{}{C}_1 \times ^{5}{}{C}_1}{^{8}{}{C}_2} = \frac{15}{28}\]
\[P\left( E_3 \right) = \frac{^{5}{}{C}_2}{^{8}{}{C}_2} = \frac{10}{28}\]
Also,
\[P\left( \frac{A}{E_1} \right) = \frac{6}{10}\]
\[P\left( \frac{A}{E_2} \right) = \frac{5}{10}\]
\[P\left( \frac{A}{E_3} \right) = \frac{4}{10}\]
∴ Required probability
= Probability that two red balls were transferred from A to B given that the ball drawn from bag B is red .
\[= P\left( \frac{E_1}{A} \right) \]
\[ = \frac{P\left( E_1 \right)P\left( \frac{A}{E_1} \right)}{P\left( E_1 \right)P\left( \frac{A}{E_1} \right) + P\left( E_2 \right)P\left( \frac{A}{E_2} \right) + P\left( E_3 \right)P\left( \frac{A}{E_3} \right)} \left[ \text { Using Baye's Theorem } \right] \]
\[ = \frac{\frac{3}{28} \times \frac{6}{10}}{\frac{3}{28} \times \frac{6}{10} + \frac{15}{28} \times \frac{5}{10} + \frac{10}{28} \times \frac{4}{10}}\]
\[ = \frac{18}{18 + 75 + 40}\]
\[ = \frac{18}{133}\]
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