Question

PLEASE HELP!!

A group of people living in either an apartment or a house are asked whether they own a pet or not. The data are collected in the table.

Answer choices in pic!!!

Answer 1

36 + 18 = 54

(1) 18/54 = 1/3 = 33%

43 + 24 = 67

(2) 24/67 = 36%




I hope these are right

Answer 2

First blank - 33%

Second blank - 36%

Explanation:

We are given the table,

Pet No Pet Total

Apartment 18 36 54

House 43 24 67

Total 61 60 121

Now, the conditional probability of an event A, given event B is P(A|B),

where
`P(A|B)=(P(A\bigcap B))/(P(B))`.

So, we have,

Probability of people living in the apartment having a pet is given by,

`P(Having\ a\ pet|Living\ in\ apartment)=(P(Having\ a\ pet\bigcap Living\ in\ apartment))/(P(Living\ in\ apartment))\\\\P(Having\ a\ pet|Living\ in\ apartment)=(18)/(54)\\\\P(Having\ a\ pet|Living\ in\ apartment)=0.33`

That is, Probability of people living in the apartment having a pet is 33%.

Also, Probability of people living in the house having no pet is given by,

`P(Having\ no\ pet|Living\ in\ house)=(P(Having\ no\ pet\bigcap Living\ in\ house))/(P(Living\ in\ house))\\\\P(Having\ no\ pet|Living\ in\ house)=(24)/(67)\\\\P(Having\ no\ pet|Living\ in\ house)=0.36`

That is, Probability of people living in the house having no pet is 36%.