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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