An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44.

Calculate the number of blue balls in the second urn.

• 4
• 20
• 24
• 44
• 64

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

Solution: A

For i = 1, 2, let [math]R_i[/math] = event that a red ball is drawn form urn i, and [math]B_i[/math] = event that a blue ball is drawn from urn i .

Then if x is the number of blue balls in urn 2,

[[math]] \begin{align*} 0.44 &= \operatorname{P}[( R1 ∩ R2 ) ∪ ( B1 ∩ B2 )] \\ &= \operatorname{P}[ R1 ∩ R2 ] + \operatorname{P}[ B1 ∩ B2 ] \\ &= \operatorname{P}[ R1 ] \operatorname{P}[ R2 ] + \operatorname{P}[ B1 ] \operatorname{P}[ B2 ] \\ &= \frac{4}{10} \frac{16}{x+16} + \frac{6}{16} \frac{x}{x+16} \end{align*} [[/math]]

Therefore,

[[math]] \begin{align*} 2.2 &= \frac{32}{x+16} + \frac{3x}{x+16} \\ 2.2 x + 35.2 &= 3 x + 32 \\ 0.8 x &= 3.2 \\ x &= 4. \end{align*} [[/math]]

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