Answer:
a.[tex]P(E_1/A)=0.0789[/tex]
b.[tex]P(E_2/A)=0.395[/tex]\
c.[tex]P(E_3/A)=0.526[/tex]
Step-by-step explanation:
Let [tex]E_1,E_2,E_3[/tex] are the events that denotes the good drive, medium drive and poor risk driver.
[tex]P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20[/tex]
Let A be the event that denotes an accident.
[tex]P(A/E_1)=0.01[/tex]
[tex]P(A/E_2=0.03[/tex]
[tex]P(A/E_3)=0.10[/tex]
The company sells Mr. Brophyan insurance policy and he has an accident.
a.We have to find the probability Mr.Brophy is a good driver
Bayes theorem,[tex]P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}[/tex]
We have to find [tex]P(E_1/A)[/tex]
Using the Bayes theorem
[tex]P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}[/tex]
Substitute the values then we get
[tex]P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}[/tex]
[tex]P(E_1/A)=0.0789[/tex]
b.We have to find the probability Mr.Brophy is a medium driver
[tex]P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395[/tex]
c.We have to find the probability Mr.Brophy is a poor driver
[tex]P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526[/tex]
