The number of severe storms that strike city J in a year follows a binomial distribution with [math]n = 5 [/math] and [math]p = 0.6 [/math]. Given that [math]m[/math] severe storms strike city J in a year, the number of severe storms that strike city K in the same year is [math]m[/math] with probability 1/2, [math]m+1[/math] with probability 1/3, and [math]m+2[/math] with probability 1/6.

Calculate the expected number of severe storms that strike city J in a year during which 5 severe storms strike city K.

• 3.5
• 3.7
• 3.9
• 4.0
• 5.7

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

Solution: C

Let J and K be the random variables for the number of severe storms in each city.

[[math]] \begin{align*} \operatorname{P}(J=j | K = 5) &= \frac{P ( K= 5 | J = j ) \operatorname{P}J=j)}{\operatorname{P}K=5)} \\ \operatorname{P}(K= 5 | J= 3) &= 1/ 6, \, \operatorname{P}J=3) = \binom{5}{3} 0.6^30.4^2 = 0.3456 \\ \operatorname{P}(K= 5 | J= 3) &= 1/3, \, \operatorname{P}J=4) = \binom{5}{4} 0.6^40.4^1 = 0.2592 \\ \operatorname{P}(K= 5 | J= 5) &= 1/ 2, \operatorname{P}J= 5)= \binom{5}{5} 0.6^50.4^0 = 0.07776 \\ \operatorname{P}(K= 5) &= (1/ 6)(0.3456) + (1/ 3)(0.2592) + (1/ 2)(0.07776) = 0.18288 \\ \operatorname{P}(J= 3 | K= 5) &= \frac{(1/ 6)(0.3456)}{0.18288} = 0.31496 \\ \operatorname{P} ( J= 4 | K= 5) &=\frac{(1/ 3)(0.2592)}{0.18288} = 0.47244 \\ \operatorname{P} ( J= 5 | K= 5) &= \frac{(1/ 2)(0.07776)}{0.18288} = 0.21260 \\ \operatorname{E}(J | K=5) &= 3(0.31496) + 4(0.47244) + 5(0.21260) = 3.89764. \end{align*} [[/math]]

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