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If A and B are two mutually exclusive events with P(A) = 0.35 and P(B) = 0.55, find the following probabilities: a. P(A ∩ B) =____ b. P(A ∪ B) =____ c. P(A)' =____ d. P(B)' =____ e. P(A ∪ B)' =____ f. P(A ∩ B' ) =____

2 Answers

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Answer:

a) P(A∩B)=0.35×0.55=0.1925

b)P(A∪B)=0.35+0.55=0.9

c)P(A)'=1-0.35=0.65

d)P(B)'=0.45

e)P(A∪B)'=0.1

f)P(A∩B)'=1-0.1925=0.8075

Explanation:

a) P(A∩B)=0.35×0.55=0.1925

b)P(A∪B)=0.35+0.55=0.9

c)P(A)'=1-0.35=0.65

d)P(B)'=0.45

e)P(A∪B)'=0.1

f)P(A∩B)'=1-0.1925=0.8075

User Zvisofer
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The probabilities are: a. P(A ∩ B) = 0; b. P(A ∪ B) = 0.90; c. P(A)' = 0.65; d. P(B)' = 0.45; e. P(A ∪ B)' = 0.10; f. P(A ∩ B') = 0.35.

How to find the probabilities?

In probability theory, if events A and B are mutually exclusive, it means that they cannot occur at the same time. The probability of the intersection of mutually exclusive events (A ∩ B) is zero.

Given:

P(A) = 0.35

P(B) = 0.55

a. P(A ∩ B) = 0 (mutually exclusive events have no intersection)

b. P(A ∪ B) = P(A) + P(B) (since A and B are mutually exclusive, we don't subtract (P(A ∩ B)):

P(A ∪ B) = 0.35 + 0.55 = 0.90

c. P(A') (the complement of A) is the probability of the complement of A occurring, i.e., the probability that A does not occur:

P(A') = 1 - P(A) = 1 - 0.35 = 0.65

d. P(B') (the complement of B) is the probability of the complement of B occurring, i.e., the probability that B does not occur:

P(B') = 1 - P(B) = 1 - 0.55 = 0.45

e. P(A ∪ B) is the complement of the union of A and B, i.e., the probability that neither A nor B occurs:

P(A ∪ B)' = 1 - P(A ∪ B) = 1 - 0.90 = 0.10

f. P(A ∩ B') is the probability of A occurring while B does not occur:

P(A ∩ B') = P(A) = 0.35

User Kubator
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