Respuesta :
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:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a person likes Italian food but not Chinese and [tex]A \cap B[/tex] is the probability that a person likes both Italian and Chinese food.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The probability that a person likes both foods is 0.21.
This means that [tex]A \cap B = 0.21[/tex]
The probability that a person likes Chinese food is 0.28
This means that [tex]B = 0.28[/tex]
So
[tex]B = b + (A \cap B)[/tex]
[tex]0.28 = b + 0.21[/tex]
[tex]b = 0.07[/tex]
The probability that a person likes Italian food is 0.75
This means that [tex]A = 0.75[/tex]
So
[tex]A = a + (A \cap B)[/tex]
[tex]0.75 = a + 0.21[/tex]
[tex]a = 0.54[/tex]
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
[tex]P = a + b + (A \cap B) = 0.54 + 0.07 + 0.21 = 0.82[/tex]
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
