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

Answer:

First blank - 33%

Second blank - 36%

Step-by-step 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 $P(A|B)=\frac{P(A\bigcap B)}{P(B)}$.

So, we have,

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

$P(Having\ a\ pet|Living\ in\ apartment)=\dfrac{P(Having\ a\ pet\bigcap Living\ in\ apartment)}{P(Living\ in\ apartment)}\\\\P(Having\ a\ pet|Living\ in\ apartment)=\dfrac{18}{54}\\\\P(Having\ a\ pet|Living\ in\ apartment)=0.33$

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,

$P(Having\ no\ pet|Living\ in\ house)=\dfrac{P(Having\ no\ pet\bigcap Living\ in\ house)}{P(Living\ in\ house)}\\\\P(Having\ no\ pet|Living\ in\ house)=\dfrac{24}{67}\\\\P(Having\ no\ pet|Living\ in\ house)=0.36$

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