$\begin{array}{1 1} \large\frac{1}{52} \\ \large\frac{3}{52} \\ \large\frac{7}{52} \\ \large\frac{21}{52} \end{array} $

- Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)

Given the total number of drivers are as follows: 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total = 2000+4000+6000 = 12000.

Let $E_1$ be the event that the insured person is a scooter driver.

P ($E_1$) = $\large\frac{\text{Number of scooter drivers}}{\text{Total number of drivers}} = \frac{2000}{12000} = \frac{1}{6}$

Let $E_2$ be the event that the insured person is a car driver.

P ($E_2$) = $\large\frac{\text{Number of car drivers}}{\text{Total number of drivers}} = \frac{4000}{12000} = \frac{1}{3}$

Let $E_3$ be the event that the insured person is a truck driver.

P ($E_3$) = $\large\frac{\text{Number of truck drivers}}{\text{Total number of drivers}} = \frac{6000}{12000} = \frac{1}{2}$

Let A: event that the insured person met w an accident.

P (scooter driver met w/ an accident) = P (A|$E_1$) = 0.01 = $\large\frac{1}{100}$

P (car driver met w/ an accident) = P (A|$E_2$) = 0.03 = $\large\frac{3}{100}$

P (truck driver met w/ an accident) = P (A|$E_3$) = 0.15 = $\large\frac{15}{100}$

The probability that a driver is a scooter driver who met w/ an accident is given by \(P(E_1/A)\).

We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)+ P(E_3)P(A|E_3)}$

P ($E_1$|A) =\(\large \frac{\frac{1}{6}.\frac{1}{100}}{\frac{1}{6} .\frac{1}{100}+\frac{1}{3}.\frac{3}{100}+\frac{1}{2}.\frac{15}{100}}\) =\(\large\frac{1}{1+6+45}\) = $\large\frac{1}{52}$

Extra: We can easily show that P (that he is a car driver) $ = \large\frac{6}{52}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...