Exercise


ABy Admin
May 01'23

Answer

Solution: D

Let D be the number of diamonds selected and S be the number of spades. First obtain the hypergeometric probability S = 0:

[[math]] \operatorname{P}(S = 0) = \frac{\binom{3}{0} \binom{7}{2}}{\binom{10}{2}} = \frac{1(21)}{45} = \frac{7}{15}. [[/math]]

The required probability distribution is:

[[math]] \operatorname{P}(D = 0 | s = 0 ) = \frac{\operatorname{P}(D = 0, S = 0)}{\operatorname{P}(S=0)} = \frac{1}{7/15} = \frac{\binom{2}{0} \binom{3}{0} \binom{5}{2}}{\binom{10}{2}} = \frac{15}{7} \frac{1(1)(10)}{45} = \frac{10}{21} [[/math]]

[[math]] \operatorname{P}(D = 1 | s = 0 ) = \frac{\operatorname{P}(D = 1, S = 0)}{\operatorname{P}(S=0)} = \frac{1}{7/15} = \frac{\binom{2}{1} \binom{3}{0} \binom{5}{1}}{\binom{10}{2}} = \frac{15}{7} \frac{2(1)(5)}{45} = \frac{10}{21} [[/math]]

[[math]] \operatorname{P}(D = 2 | s = 0 ) = \frac{\operatorname{P}(D = 2, S = 0)}{\operatorname{P}(S=0)} = \frac{1}{7/15} = \frac{\binom{2}{2} \binom{3}{0} \binom{5}{0}}{\binom{10}{2}} = \frac{15}{7} \frac{1(1)(1)}{45} = \frac{1}{21} [[/math]]

Then

[[math]] \begin{align*} E(D | S =0) = 0(10 / 21) + 1(10 / 21) + 2(1/ 21) = 12 / 21 = 4/7 \\ E(D^2 | S = 0) = 0^2 (10/21) + 1^2(10/21) + 2^2(1/21) = 14/21 = 2/3 \\ \operatorname{Var}(D | S = 0 ) = 2/3 - (4/7)^2 = 50/147 = 0.34. \end{align*} [[/math]]

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

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