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)}=((22+42))/(162)=(64)/(162)=0.395`
`P(\text{Own pet)=}((54+40))/(162)=(94)/(162)=0.580`
`\begin{gathered} P(Male|Own\text{ pet)=}\frac{P(\text{Male }\cap\text{ Own pet)}}{P(Own\text{ pet)}} \\ =(54)/((54+40))=(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)}} \\ =(22)/(62)=0.355 \end{gathered}`

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