An insurance company insured 2000 light vehicle drivers, 4000 medium vehicle drivers and 6000 heavy vehicle drivers.The probability of them meeting an accident are 0.01, 0.03 and 0.15 respectively. If one of the insured drivers meets with an accident what is the probability that he/she was a medium vehicle driver.

Concept:

Bayes' Theorem:

Let E1, E2, ….., En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or … or En such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

Calculation:

Let the events B1, B2 and B3 be defined as:

B1 : The insured person is light vehicle driver

B2 : The insured person is medium vehicle driver

B3 : The insured person is heavy vehicle driver

Let the event E be 'The insured person meeting an accident'

⇒P (B₁) = 2000/12000 = 1/6, P  (B₂) = 4000/12000 = 1/3 and P (B₃) = 6000/12000 = 1/2

Similarly, P (E | B1) = 0.01, P (E | B2) = 0.03 and P (E | B3) = 0.15 ------(Given)

Here, we have to find the value of P (B₂ | E)

As we know that according to bayes' theorem: \(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)

\( \Rightarrow P{\rm{(}}{B_2}\;{\rm{|}}E) = \frac{{P\left( {{B_2}} \right) \cdot P\;\left( {E\;|\;{B_2}} \right)}}{{\left[ {P\left( {{B_1}} \right) \cdot P\;\left( {E\;|\;{B_1}} \right) + \;P\left( {{B_2}} \right) \cdot P\;\left( {E\;|\;{B_2}} \right) + P\left( {{B_3}} \right) \cdot P\;\left( {E\;|\;{B_3}} \right)} \right]}}\;\)

\( \Rightarrow P{\rm{(}}{B_2}\;{\rm{|}}E) = \frac{3}{{26}}\)