Bowl I contains eight red balls and six blue balls. Bowl II is empty. Four balls are selected at random, without replacement, and transferred from bowl I to bowl II. One ball is then selected at random from bowl II.

Calculate the conditional probability that two red balls and two blue balls were transferred from bowl I to bowl II, given that the ball selected from bowl II is blue.

• 0.21
• 0.24
• 0.43
• 0.49
• 0.57

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

[[math]] \begin{align*} \operatorname{P}( \textrm{2 red and 2 blue transferred} | \textrm{blue drawn} ) = \frac{\operatorname{P}( \textrm{2 red and 2 blue transferred and blue drawn})}{\operatorname{P}(\textrm{blue drawn})} \\ \operatorname{P}( \textrm{2 red and 2 blue transferred and blue drawn}) = \frac{\binom{8}{2} \binom{6}{2}}{\binom{14}{4}} \times \frac{2}{4} = \frac{840}{4004} \\ \operatorname{P}( \textrm{blue drawn} ) = \frac{\binom{8}{0} \binom{6}{4}}{\binom{14}{4}} \times \frac{4}{4} + \frac{\binom{8}{1} \binom{6}{3}}{\binom{14}{4}} \times \frac{3}{4} + \frac{\binom{8}{2}\binom{6}{2}}{\binom{14}{4}} \times \frac{1}{4} = \frac{60 + 480 + 840 + 336}{4004} = \frac{1716}{4004} \\ \operatorname{P}(\textrm{2 red and blue transferred} | \textrm{blue drawn} ) = \frac{840}{1716} = 0.49. \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.