Revision as of 18:52, 29 April 2023 by Admin (Created page with "An actuary studied the likelihood that different types of drivers would be involved in at least one collision during any one-year period. The results of the study are: {| cla...")
ABy Admin
Apr 29'23
Exercise
An actuary studied the likelihood that different types of drivers would be involved in at least one collision during any one-year period. The results of the study are:
Type of Driver | Percentage of all drivers | Probability of at least one collision |
---|---|---|
Teen | 8% | 0.15 |
Young adult | 16% | 0.08 |
Midlife | 45% | 0.04 |
Senior | 31% | 0.05 |
Given that a driver has been involved in at least one collision in the past year, calculate the probability that the driver is a young adult driver.
- 0.06
- 0.16
- 0.19
- 0.22
- 0.25
ABy Admin
Apr 29'23
Solution: D
Let
[math]C[/math] = Event of a collision
[math]T[/math] = Event of a teen driver
[math]Y[/math] = Event of a young adult driver
[math]M[/math] = Event of a midlife driver
[math]S[/math] = Event of a senior driver
Then using Bayes’ Theorem, we see that
[[math]]
\begin{align*}
\operatorname{P}[Y|C] &= \frac{\operatorname{P}[C | Y ]\operatorname{P}[Y ]}{\operatorname{P}[C | T ]\operatorname{P}[T ] + \operatorname{P}[C | Y ]\operatorname{P}[Y ] + \operatorname{P}[C | M ]\operatorname{P}[ M ] + \operatorname{P}[C | S ]\operatorname{P}[ S ]} \\
&= \frac{(0.08)(0.16)}{(0.15)(0.08) + (0.08)(0.16) + (0.04)(0.45) + (0.05)(0.31)} \\
&= 0.22
\end{align*}
[[/math]]