Question

Events A and B are independent. Suppose event occurs with probability.89 and event

occurs with probability .16.
Compute the following. (If necessary, consult a list of formulas.)
(a) Compute the probability that A does not occur or B does not occur (or both).

(b) Compute the probability that either B occurs without A occurring or A and B
both occur.

Answer 1

Answers:

(a) 0.8576

(b) 0.16

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Work Shown for Part (a)

P(A) = 0.89

P(not A) = 1 - P(A)

P(not A) = 1 - 0.89

P(not A) = 0.11

P(B) = 0.16

P(not B) = 1 - P(B)

P(not B) = 1 - 0.16

P(not B) = 0.84

P(not A and not B) = P(not A)*P(not B)

P(not A and not B) = 0.11*0.84

P(not A and not B) = 0.0924

P(not A or not B) = P(not A) + P(not B) - P(not A and not B)

P(not A or not B) = 0.11 + 0.84 - 0.0924

P(not A or not B) = 0.8576

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An alternative approach to part (a)

P(A) = 0.89

P(B) = 0.16

P(A and B) = P(A)*P(B)

P(A and B) = 0.89*0.16

P(A and B) = 0.1424

P(not A or not B) = P( not (A and B) )

P(not A or not B) = 1 - P(A and B)

P(not A or not B) = 1 - 0.1424

P(not A or not B) = 0.8576

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Work Shown for part (b)

P(A) = 0.89

P(B) = 0.16

P(not A) = 1 - P(A)

P(not A) = 1 - 0.89

P(not A) = 0.11

P(B and not A) = P(B)*P(not A)

P(B and not A) = 0.16*0.11

P(B and not A) = 0.0176

P(B and A) = P(B)*P(A)

P(B and A) = 0.16*0.89

P(B and A) = 0.1424

P( (B and not A) or (B and A) ) = P(B and not A) + P(B and A)

P( (B and not A) or (B and A) ) = 0.0176 + 0.1424

P( (B and not A) or (B and A) ) = 0.16

It is not a coincidence that we end up with P(B) = 0.16

In this case, we're asking about the probability of B occurring regardless if A occurs or not. So we're just looking to compute P(B).

Refer to the law of total probability for more information.