Q 3.50 - A First Course in Probability

Question

 Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company’s 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 is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder A had no accidents in 2012, what is the probability that he or she is a good risk? is an average risk? 

Step-by-Step Solution

Answer

With events G, M, B that a person is in the good, average or bad category, and A event that accident happened in a fixed year

$\left[2\mathrm{ex}\right]P\left(A\right)=0.175$

The probability that he or she is a good risk an average risk is

$P\left(G\cup M\mid A^c\right)\approx 0.7455$

Step 1:Given Information

Events:

G - a person is in the good risk category

M-a person is in the average risk category

B - a person is in the bad risk category

A - a person has an accident in a fixed year

Probabilities:$P\left(G\right)=20\%=0.2 P\left(M\right)=50\%=0.5 P\left(B\right)=30\%=0.3$

Since the boxes' contents are known:

$P(A \backslash G)=0.05 P(A \backslash M)=0.15 P(A \backslash B)=0.30$

Calculate: $\left.P\left(A\right),P\left(G\cup M\mid A^c\right)\right\}$

P(A) can be obtained by conditioning on whether a person is a good, an average or a bad risk:

$P\left(A\right)=P(A\mid G)P\left(G\right)+P(A\mid M)P\left(M\right)+P(A\mid B)P\left(B\right)$

By substituting before stated probabilities the result is :

$P\left(A\right)=0.175$

Step 2:Calculation

Conditional probability is also a probability ,and G and M are mutually exclusive, so by Axiom 3:

$P\left(G\cup M\mid A^c\right)=P\left(G\mid A^c\right)+P\left(M\mid A^c\right)$

For $P\left(G\mid A^c\right)$ use the definition formula for conditional probability:

   $P\left(G\mid A^c\right)=\frac{P\left(G{A}^c\right)}{P\left(A^c\right)}$

As$P\left(A\right)=0.175$ is already calculated, a proposition states :

 $P\left(A^c\right)=1-P\left(A\right)\Rightarrow P\left(A^c\right)=0.825$

And by a different use of the definition of $P\left(A^c\mid G\right)$:

   $P\left(G{A}^C\right)=P\left(A^C/G\right) P\left(G\right)$

Conditional probability is also a probability so:

$P\left(A^C/G\right)=1-P\left(A/G\right)\Rightarrow P\left(A^C/G\right)=0.95$

Now all the elements of the equation are known, so we can calculate :

    $P\left(G/A^C\right)=\frac{0.95\times 0.2}{0.825}\approx 0.2303$

The same is applied for $P\left(M/A^C\right)$:

$P\left(M/A^C\right)=\frac{P\left(M{A}^C\right)}{P\left(\left(A^C\right)\right)}$

$P\left(A^{C^{}}\right)=0.825$

 $$\begin{gathered} P\left(M{A}^C\right)=P\left(A^C/M\right) P\left(M\right), P\left(A^{C}M\right)=1-P\left(A/M\right) \\ \Rightarrow P\left(M{A}^C\right)=0.85\times 0.5=0.425 \\ P\left(G/A^C\right)=\frac{0.425}{0.825}\approx 0.5152 \end{gathered}$$

Step 3 :Final Answer

Conditional probability is also a probability so:

$P\left(G\cup M/A^C\right)\approx 0.7455$