Probability Theory

alakaboom1

New member
An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?

I think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesn't account for the fact that we are assuming that the driver is already in an accident...

galactus

Super Moderator
Staff member
You are looking for the probability the driver is from class A given they have an accident:

$$\displaystyle P(\text{class A}|\text{accident})$$

You can make a chart. Assume a number of drivers that is easy to work with. Say, 1000

Or, use Bayes Theorem. Do you know it?.

soroban

Elite Member
Hello, alakaboom1!

Here is a very primitive approach . . .

An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C.
The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%.
The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively.
After purchasing an insurance policy, a driver has an accident within the first year.
What is the probability that the driver is of class A?

Suppose there are 10,000 policyholders.

$$\displaystyle \text{20\% are class A drivers: }\:20\%\times 10,\!000 \,=\,2000$$
. . $$\displaystyle \text{1\% of them will have an accident: }\:1\% \times 2000 \,=\,20$$

$$\displaystyle \text{65\% are class B drivers: }\:65\% \times 10,\!000 \,=\,6500$$
. . $$\displaystyle \text{2\% of them will have an accident: }\:2\% \times 6500 \,=\,130$$

$$\displaystyle \text{15\% are class C drivers: }\:15\% \times 10,\!000 \,=\,1500$$
. . $$\displaystyle \text{3\% of them will have an accident: }\:3\% \times 1500 \,=\,45$$

$$\displaystyle \text{There is a total of: }\:20 + 130 + 45 \,=\,195\text{ accidents}$$
. . $$\displaystyle \text{of which 20 are class A drivers.}$$

$$\displaystyle \text{Therefore: }\(\text{class A}\,|\,\text{accident}) \;=\;\frac{20}{195} \;=\;\frac{4}{39}$$

galactus

Super Moderator
Staff member
That 'primitive
approach is a nice explanation.

Here is all I was getting at:

$$\displaystyle \frac{(.01)(.2)}{(.01)(.2)+(.02)(.65)+(.03)(.15)}=\frac{4}{39}\approx .1025$$

When confronted with a 'probability of something given something' problem, you can make a chart.

Assume 1000 policyholders.

$$\displaystyle \begin{array}{c|c|c|c|c}\text{}&A&B&C&\text{total}\\ \hline\text{has accidents}&2&13&4.5&19.5 \\ \hline\text{no accident}&198&637&145.5&980.5 \\ \hline\text{total}&200&650&150&1000\end{array}$$

Now, you can answer any problem they throw at you. In this case, we want

"Probability the driver is class A given they have an accident".

Go down the class A column and across the 'has accident' row. $$\displaystyle \frac{2}{19.5}=\frac{4}{39}$$

Say they asked for "probability the driver is from class C given they did not have an accident?"

Go down the C column and across the 'no accident' row and get $$\displaystyle \frac{145.5}{980.5}=\frac{291}{1961}\approx .148$$

and so on..............

alakaboom1

New member
Thanks a lot guys!

Out of curiosity, is this formula you guys mentioned known as Bayes Theorem? I've definitely used the formula before, but never knew it by name.