Answer:
a. 17.5 %
b. 0.2303
Step-by-step explanation:
Let's start defining the conditional probability :
Suppose two events A and B where [tex]P(A)>0[/tex] and [tex]P(B)>0[/tex]
and P(A ∩ B) = P(A,B) where (A ∩ B) is the event where A and B occur both at the same time.
The conditional probability :
[tex]P(A/B)=\frac{P(A,B)}{P(B)} \\P(B/A)=\frac{P(A,B)}{P(A)}[/tex]
Let's define the following events :
A : ''Randomly chosen person had accident in a fixed year''
GR : ''The person belongs to good risks classification''
AR : ''The person belongs to average risks classification''
BR : ''The person belongs to bad risks classification''
The information given is :
[tex]P(GR)=0.20\\P(AR)=0.50\\P(BR)=0.30[/tex]
[tex]P(A/GR)=0.05\\P(A/AR)=0.15\\P(A/BR)=0.30[/tex]
a.
We need to calculate [tex]P(A)[/tex]
[tex]P(A)=P(A,GR)+P(A,AR)+P(A,BR)\\P(A)=P(A/GR).P(GR)+P(A/AR).P(AR)+P(A/BR).P(BR)\\P(A)=(0.05)(0.2)+(0.15)(0.5)+(0.30)(0.30)\\P(A)=0.175[/tex]
Then 17.5% of people have accidents in a fixed year
b. If U is an event ⇒
[tex]P(U)=1-P(U^{c})[/tex]
Where [tex]U^{c}[/tex] is the event where U does not occur
We need to calculate :
[tex]P(GR/A^{c})[/tex]
[tex]P(A^{c})=1-P(A)=1-0.175\\P(A^{c})=0.825[/tex]
[tex]P(A^{c}/GR)=1-P(A/GR)=1-0.05\\P(A^{c}/GR)=0.95[/tex]
[tex]P(A^{c}/GR)=\frac{P(A^{c},GR)}{P(GR)} \\0.95=\frac{P(A^{c},GR)}{0.2} \\P(A^{c},GR)=(0.95)(0.2) \\P(A^{c},GR)=0.19[/tex]
[tex]P(GR/A^{c})=\frac{P(GR,A^{c})}{P(A^{c})}=\frac{0.19}{0.825}\\P(GR/A^{c})=0.2303[/tex]
