excans:9b011a2d0c: Difference between revisions
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(Created page with "'''Solution: C''' Let x be the probability of having all three risk factors. <math display = "block"> \frac{1}{3} = \operatorname{P}[A \cap B \cap C | A \cap B ] = \frac{\op...") |
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<math display = "block"> | <math display = "block"> | ||
\begin{align*} | \begin{align*} | ||
\operatorname{P}[( A ∪ B ∪ C ) | A^c] &= \frac{\operatorname{P}[( A ∪ B ∪ C )]}{\operatorname{P}[A^c]} \\ | \operatorname{P}[( A ∪ B ∪ C )^c | A^c] &= \frac{\operatorname{P}[( A ∪ B ∪ C )^c]}{\operatorname{P}[A^c]} \\ | ||
&= \frac{1− \operatorname{P}[ A ∪ B ∪ C]}{1-\operatorname{P}[A]} \\ | &= \frac{1− \operatorname{P}[ A ∪ B ∪ C]}{1-\operatorname{P}[A]} \\ | ||
&= \frac{1 − 3 ( 0.10 ) − 3 ( 0.12 ) − 0.06}{1 − 0.10 − 2 ( 0.12 ) − 0.06} \\ | &= \frac{1 − 3 ( 0.10 ) − 3 ( 0.12 ) − 0.06}{1 − 0.10 − 2 ( 0.12 ) − 0.06} \\ | ||
Latest revision as of 22:19, 1 July 2024
Solution: C
Let x be the probability of having all three risk factors.
[[math]]
\frac{1}{3} = \operatorname{P}[A \cap B \cap C | A \cap B ] = \frac{\operatorname{P}[A \cap B \cap C]}{\operatorname{P}[A \cap B]}
[[/math]]
It follows that
[[math]]
\begin{align*}
x = \frac{1}{3}(x + 0.12 ) &= \frac{1}{3}x + 0.04 \\
\frac{2}{3}x &= 0.04 \\
x &= 0.06.
\end{align*}
[[/math]]
Now we want to find
[[math]]
\begin{align*}
\operatorname{P}[( A ∪ B ∪ C )^c | A^c] &= \frac{\operatorname{P}[( A ∪ B ∪ C )^c]}{\operatorname{P}[A^c]} \\
&= \frac{1− \operatorname{P}[ A ∪ B ∪ C]}{1-\operatorname{P}[A]} \\
&= \frac{1 − 3 ( 0.10 ) − 3 ( 0.12 ) − 0.06}{1 − 0.10 − 2 ( 0.12 ) − 0.06} \\
&= \frac{0.28}{0.60} = 0.467
\end{align*}
[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.