Question

A local university wanted to understand what students prefer to eat during finals. They asked 1000 students, "Do you prefer chicken, burgers, or pizza?" The results of the survey are shown in the two-way table below:
Chicken Burgers Pizza
Male 127 218 125
Female 219 192 119


What is the probability that a person chosen at random from this survey prefers pizza given that they are female? Round your answer to the nearest tenth.

Answer 1

Answer:

$P(\text{Pizza}\ |\ \text{Female})=0.225$

Step-by-step explanation:

The two-way frequency table is attached below.

We have to calculate the probability of, a person chosen at random prefers pizza given that they are female, i.e $P(\text{Pizza}\ |\ \text{female})$

This is a conditional probability.

We know that,

$P(A\ |\ B)=\dfrac{P(A\ \cap\ B)}{P(B)}$

So,

$P(\text{Pizza}\ |\ \text{Female})=\dfrac{P(\text{Pizza}\ \cap\ \text{Female)}}{P({\text{Female}})}$

From the table,

$P(\text{Pizza}\ \cap\ \text{Female)}=\dfrac{119}{1000}$

$P({\text{Female}})=\dfrac{530}{1000}$

Putting the values,

$P(\text{Pizza}\ |\ \text{Female})=\dfrac{P(\text{Pizza}\ \cap\ \text{Female)}}{P({\text{Female}})}=\dfrac{\frac{119}{1000}}{\frac{530}{1000}}=0.225$