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Given P(A) = 0.74 , P(B) = 0.25 , and P(A ∩B) = 0.13. Calculate
(1) P(A ∪B)
(2) P(A ∩B’)
(3) P(A’|B’)
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(1) It is easy.
Use the basic general formula of elementary probability theory
P(A U B) = P(A) + P(B) - P(A ∩ B) = 0.74 + 0.25 - 0.13 = 0.86. ANSWER
(2) To calculate P(A ∩ B’), ask yourself what is the set (A ∩ B’) ?
This set are those elements of A that do not belong to B.
In other words, (A ∩ B’) = A \ (A ∩ B), where the sign " \ " means subtracting the set (A ∩ B) from A.
After this notice, it is OBVIOUS that P(A ∩ B’) = P(A) - P(A ∩ B) = 0.74 - 0.13 = 0.61. ANSWER
(3) P(A'|B') is the conditional probability P(A' ∩ B’) / P(B').
To calculate P(A' ∩ B’), ask yourself what is the set (A' ∩ B’) ?
It is the set of elements that do not belong NEITHER A NOR B.
In other words, it is the COMPLEMENT of the set (A U B) to the universal set.
THEREFORE, P(A' ∩ B’) = 1 - P(A U B) = 1 - 0.86 = 0.14,
as we just defined the value of P(A U B) = 0.86 in n.(1) above.
After this notice, it is OBVIOUS that P(A'|B') = P(A' ∩ B’) / P(B') = = = = 0.186667 (rounded). ANSWER
Solved. // All questions are answered.
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To see other similar solved problems, look into the lesson
- Using general probability formulas for a union or intersection of events
in this site.
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