Question

A random sample of a 164 students were active they own a pet or not the fallen contingency table gives the 2 way classification of their response

Answer 1

Given a random sample of 164 students and a contingency table which gives the 2 way classification of their responses.

The probability formula is:

$P(E)=\frac{n(\text{ Required outcome )}}{n\text{ (Possible outcome)}}$$P(\text{Female)}=\frac{(22+42)}{162}=\frac{64}{162}=0.395$$P(\text{Own pet)=}\frac{(54+40)}{162}=\frac{94}{162}=0.580$$\begin{gathered} P(Male|Own\text{ pet)=}\frac{P(\text{Male }\cap\text{ Own pet)}}{P(Own\text{ pet)}} \\ =\frac{54}{(54+40)}=\frac{54}{94}=0.574 \end{gathered}$$\begin{gathered} P(No\text{ pet}|\text{Female)}=\frac{P(No\text{ pet }\cap Female\text{)}}{P(\text{Female)}} \\ =\frac{22}{62}=0.355 \end{gathered}$

Note, all answers are rounded to 3 decimal places as required