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,
$$\begin{gathered} \mathrm{P}\left(E_1\right)=\frac{{}^{3}C_2}{{}^{8}C_2}=\frac{3}{28} \\ \mathrm{P}\left(E_2\right)=\frac{{}^{3}C_1\times {}^{5}C_1}{{}^{8}C_2}=\frac{15}{28} \\ \mathrm{P}\left(E_3\right)=\frac{{}^{5}C_2}{{}^{8}C_2}=\frac{10}{28} \end{gathered}$$
Also,
$$\begin{gathered} \mathrm{P}\left(\frac{A}{E_{\mathit{1}}}\right)=\frac{6}{10} \\ \mathrm{P}\left(\frac{A}{E_2}\right)=\frac{5}{10} \\ \mathrm{P}\left(\frac{A}{E_3}\right)=\frac{4}{10} \end{gathered}$$
∴ Required probability

= Probability that two red balls were transferred from A to B given that the ball drawn from bag B is red

$$\begin{gathered} =\mathrm{P}\left(\frac{E_1}{A}\right) \\ =\frac{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{A}{E_1}}\right)}{\mathrm{P}\left(E_1\right)\mathrm{P}\left({\displaystyle \frac{A}{E_1}}\right)+\mathrm{P}\left(E_2\right)\mathrm{P}\left({\displaystyle \frac{A}{E_2}}\right)+\mathrm{P}\left(E_3\right)\mathrm{P}\left({\displaystyle \frac{A}{E_3}}\right)}\left[\mathrm{Using}\mathrm{Baye}\text{'}\mathrm{s}\mathrm{Theorem}\right] \\ =\frac{{\displaystyle \frac{3}{28}\times \frac{6}{10}}}{{\displaystyle \frac{3}{28}\times \frac{6}{10}+\frac{15}{28}\times \frac{5}{10}+\frac{10}{28}\times \frac{4}{10}}} \\ =\frac{18}{18+75+40} \\ =\frac{18}{133} \end{gathered}$$