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A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8. Compute the following.a. P(A ∩ B).b. The probability that the lifeline usage amount is exceeded in exactly one of the two months.

Respuesta :

Complete question is;

A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8. Compute the following.

a. P(A ∩ B).

b. The probability that the lifeline usage amount is exceeded in exactly one of the two months.

Answer:

A) 0.4

B) 0.4

Step-by-step explanation:

We are given;

P(A) = 0.7, P(B) = 0.5, and P(A ∪ B) = 0.8

A) To solve this question, we will use the the general probability addition rule for the union of two events which is;

P(A∪B) = P(A) + P(B) − P(A∩B)

Making P(A∩B) the subject of the equation, we have;

P(A∩B) = P(A) + P(B) − P(A∪B)

Thus, plugging in the relevant values, we have;

P(A∩B) = 0.7 + 0.5 - 0.8

P(A∩B) = 0.4

B)The probability that the lifeline usage amount is exceeded in exactly one of the two months can be described in terms of A and B as:

P(A but not B) + P(B but not A) = P(A∩B') + P(B∩A')

where;

A' is compliment of set A

B' is compliment of set B

Now,

P(A∩B') = 0.7 − 0.4 = 0.3

P(B∩A') = 0.5 − 0.4 = 0.1

Thus;

P(A but not B) + P(B but not A) = 0.1 + 0.3 = 0.4