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.

Answer 1

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 .

\[= P\left( \frac{E_1}{A} \right) \]

\[ = \frac{P\left( E_1 \right)P\left( \frac{A}{E_1} \right)}{P\left( E_1 \right)P\left( \frac{A}{E_1} \right) + P\left( E_2 \right)P\left( \frac{A}{E_2} \right) + P\left( E_3 \right)P\left( \frac{A}{E_3} \right)} \left[ \text { Using Baye's Theorem } \right] \]

\[ = \frac{\frac{3}{28} \times \frac{6}{10}}{\frac{3}{28} \times \frac{6}{10} + \frac{15}{28} \times \frac{5}{10} + \frac{10}{28} \times \frac{4}{10}}\]

\[ = \frac{18}{18 + 75 + 40}\]

\[ = \frac{18}{133}\]