## Related questions with answers

Suppose that an insurance company classifies people into one of three classes — good risks, average risks, and bad risks. Their records indicate that the probabilities that good, average, and bad risk persons will be involved in an accident over a 1-year span are, respectively, .05, .15, and .30. If 20 percent of the population are “good risks,” 50 percent are “average risks,” and 30 percent are “bad risks,” what proportion of people have accidents in a fixed year? If policy holder A had no accidents in 1987, what is the probability that he or she is a good (average) risk?

Solution

VerifiedDefine:

$B$ : event that a person has an accident

$B'$ : event that a person has no accident

$A_1$ : event that a person is good risk

$A_2$ : event that a person is average risk

$A_3$ : event that a person is bad risk

Now, $P(A_1)=0.20$, $P(A_2)=0.50$, $P(A_3)=0.30$, $P(B|A_1)=0.05$, $P(B|A_1)=0.15$ and $P(B|A_3)=0.30$

$\begin{align*} P(B)&=P(A_1)P(B|A_1)+P(A_2)P(B|A_2)+P(A_3)P(B|A_3)\\ &=0.20 \times 0.05+0.50 \times 0.15+0.30 \times 0.30\\ &=0.01+0.075+0.09\\ &=0.175 \end{align*}$

So,

$P(B')=1-P(B)=1-0.175=0.825$

$\begin{align*} P(A_1|B')&=\frac{P(A_1)P(B'|A_1)}{P(B')}\\ &=\frac{0.20 \times (1-0.05)}{0.825}\\ &=\frac{0.19}{0.825}\\ &=0.2303 \end{align*}$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Introduction to Probability and Statistics for Engineers and Scientists

5th Edition•ISBN: 9780123948113Sheldon Ross#### Probability and Statistics for Engineers and Scientists

9th Edition•ISBN: 9780321629111 (9 more)Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers#### Probability and Statistics for Engineers and Scientists

4th Edition•ISBN: 9781111827045 (2 more)Anthony J. Hayter#### Applied Statistics and Probability for Engineers

6th Edition•ISBN: 9781118539712 (1 more)Douglas C. Montgomery, George C. Runger## More related questions

1/4

1/7