excans:C354ffb55a: Difference between revisions
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(Created page with "'''Solution: A''' Let A = event that person wants life policy P B = event that person wants life policy Q C = event that person wants the health policy and let a, b, c, d...") |
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and let a, b, c, d be the probabilities of the regions as shown. | and let a, b, c, d be the probabilities of the regions as shown. | ||
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[[File:Examp2023188.jpg | 300px]] | |||
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#is reflected by no intersection of A and B | #is reflected by no intersection of A and B | ||
#is reflected by the 0.18 in the diagram | #is reflected by the 0.18 in the diagram | ||
Latest revision as of 12:10, 30 April 2023
Solution: A
Let A = event that person wants life policy P
B = event that person wants life policy Q
C = event that person wants the health policy
and let a, b, c, d be the probabilities of the regions as shown.
- is reflected by no intersection of A and B
- is reflected by the 0.18 in the diagram
- implies a + b = 2(c + d)
- implies b + c + 0.18 = 0.45 or b + c = 0.27
- implies P([A or B] and C) = P(A or B)P(C) or b + c = (a + b + c + d)(0.45)
So 0.27 = (a + d + 0.27)(0.45) and then a + d = 0.33. The desired probability is a + 0.18 + d = 0.33 + 0.18 = 0.51.
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.