SOLUTION: If A and B are independent events, P(A) = 0.4, and P(B) = 0.6, find the probabilities below. (a) P(A intersection B) (b) P(A union B)

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Question 355589: If A and B are independent events, P(A) = 0.4, and P(B) = 0.6, find the probabilities below.
(a) P(A intersection B)
(b) P(A union B)
Found 2 solutions by nyc_function, jim_thompson5910:
Answer by nyc_function(2741) (Show Source):
You can put this solution on YOUR website!
P(A intersection B) = probability involving AND.

P(A intersection B) = P(A) + P(B)

P(A intersection B) = 0.4 + 0.6

P(A intersection B) = 1.0 or simply 1.

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P(A union B) = probability involving OR.

P(A union B) = P(A) + P(B) - P(A intersection B)

P(A union B) = 0.4 + 0.6 - 1.0

P(A union B) = 1.0 - 1.0

P(A union B) = 0

I hope this helps.

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

Answers
(a) 0.24
(b) 0.76

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Work Shown

Part (a)

P(A intersection B) = P(A)*P(B) ... only works if A and B are independent

P(A intersection B) = 0.4*0.6

P(A intersection B) = 0.24

Note: the other tutor seemed to have mixed up some formulas. Perhaps they were thinking about P(A union B) = P(A)+P(B). This formula only works if A and B are disjoint or mutually exclusive.

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Part (b)

P(A union B) = P(A) + P(B) - P(A intersection B)

P(A union B) = 0.4 + 0.6 - 0.24

P(A union B) = 0.76