Carcass wrote:
Let A, B, C, and D be events for which P(A or B) = 0.6, P(A) = 0.2, P(C or D) = 0.6, P(C) = 0.5
The events A and B are mutually exclusive, and the events C and D are independent.
(a) Find P(B)
(b) Find P(D)
Math Review
Question: 14
Page: 298
Difficulty: medium
Key concepts:
- P(X or Y) = P(X) + P(Y) - P(X and Y)
- If X and Y are mutually exclusive events, then P(X and Y) = 0
- If X and Y are independent events, then P(X and Y) = P(X) * P(Y)
(a) Find P(B) Since events A and B are mutually exclusive, we know that P(A and B) = 0
Take: P(A or B) = P(A) + P(B) - P(A and B) and replace parts with their given values.
We get: 0.6 = 0.2 + P(B) - 0
Solve to get: P(B) = 0.4
Answer: 0.4------------------------------------------------
(b) Find P(D)Since events C and D are mutually exclusive, we know that P(C and D) = P(C) x P(D)
Take: P(C or D) = P(C) + P(D) - P(C and D) and replace parts with their given values.
We get: 0.6 = 0.5 +
P(D) - [0.5 x
P(D)]
Subtract 0.5 from both sides to get: 0.1 =
P(D) - [0.5 x
P(D)]
Factor out
P(D) to get: 0.1 =
P(D)[1 - 0.5]
Simplify to get: 0.1 =
P(D)[0.5]
Divide both sides by 0.5 to get:
P(D) = 0.1/0.5 = 1/5 = 0.2
Answer: 0.2Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com
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