Revision as of 18:59, 29 April 2023 by Admin (Created page with "'''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...")
Exercise
ABy Admin
Apr 29'23
Answer
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]]