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A random sample of 269 student were asked what kind of vehicle they prefer a car truck the following contingency table gives the 2 way classification of the responses

User Dalcantara
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Given a random sample of 269 students and a contigency table that gives the 2 way classification of their responses.

The probability formula is:


P(E)=\frac{n(\text{Required outcome)}}{n(\text{Possible outcome)}}
P(\text{Female)}=((99+22))/(269)=(121)/(269)=0.450
P(\text{Car)}=((87+99))/(269)=(186)/(269)=0.691
\begin{gathered} P(\text{Female}|\text{Truck)}=\frac{P(\text{Female }\cap\text{ Truck)}}{P(\text{Truck)}} \\ =(22)/((61+22))=(22)/(83)=0.265 \end{gathered}
P(\text{Truck }\cap\text{ Female)=}(22)/((99+22))=(22)/(121)=0.182

The conditional probability P(A/B) or P(B/A) arises only in the case of dependent events.

User Zehrer
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