Given a random sample of 269 students and a contigency table that gives the 2 way classification of their responses.
The probability formula is:
[tex]P(E)=\frac{n(\text{Required outcome)}}{n(\text{Possible outcome)}}[/tex][tex]P(\text{Female)}=\frac{(99+22)}{269}=\frac{121}{269}=0.450[/tex][tex]P(\text{Car)}=\frac{(87+99)}{269}=\frac{186}{269}=0.691[/tex][tex]\begin{gathered} P(\text{Female}|\text{Truck)}=\frac{P(\text{Female }\cap\text{ Truck)}}{P(\text{Truck)}} \\ =\frac{22}{(61+22)}=\frac{22}{83}=0.265 \end{gathered}[/tex][tex]P(\text{Truck }\cap\text{ Female)=}\frac{22}{(99+22)}=\frac{22}{121}=0.182[/tex]The conditional probability P(A/B) or P(B/A) arises only in the case of dependent events.
