Question

A survey was taken among a group of people. The probability that a person chosen likes Italian food is 0.75, the probability that a person likes Chinese food is 0.28,
and the probability that a person likes both foods is 0.21.
Determine the probability that a person likes Italian, but not Chinese
Determine the probaility that a person likes at least one of these foods
Determine the proability that a person likes at most one of these foods -

Answer 1

Answer:

54% probability that a person likes Italian food, but not Chinese food.

82% probaility that a person likes at least one of these foods

79% proability that a person likes at most one of these foods

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a person likes Italian food.

B is the probability that a person likes Chinese food.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a person likes Italian food but not Chinese and $A \cap B$ is the probability that a person likes both Italian and Chinese food.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that a person likes both foods is 0.21.

This means that $A \cap B = 0.21$

The probability that a person likes Chinese food is 0.28

This means that $B = 0.28$

So

$B = b + (A \cap B)$

$0.28 = b + 0.21$

$b = 0.07$

The probability that a person likes Italian food is 0.75

This means that $A = 0.75$

So

$A = a + (A \cap B)$

$0.75 = a + 0.21$

$a = 0.54$

Determine the probability that a person likes Italian, but not Chinese

This is a.

54% probability that a person likes Italian food, but not Chinese food.

Determine the probaility that a person likes at least one of these foods

$P = a + b + (A \cap B) = 0.54 + 0.07 + 0.21 = 0.82$

82% probaility that a person likes at least one of these foods

Determine the proability that a person likes at most one of these foods

Either a person likes at most one of these foods, or it likes both. The sum of the probabilities of these events is decimal 1.

0.21 probability it likes both.

Then

0.21 + p = 1

p = 0.79

79% proability that a person likes at most one of these foods