Revision as of 23:39, 3 May 2023 by Admin (Created page with "'''Solution: C''' Let J and K be the random variables for the number of severe storms in each city. <math display = "block"> \begin{align*} \operatorname{P}(J=j | K = 5) &=...")
Exercise
ABy Admin
May 04'23
Answer
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]]