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ladessa [460]
3 years ago
7

A survey was taken among a group of people. The probability that a person chosen likes Italian food is 0.75, the probability tha

t 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 -
Mathematics
1 answer:
Luda [366]3 years ago
6 0

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

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