A smoke-detector system uses two devices, A and B. If smoke is present, the probability that...

A smoke-detector system uses two devices, A and B. If smoke is present, the probability that it will be detected by device A is 0.95; by device B is 0.98; and both devices, 0.94. Suppose smoke is present in a home that has the above smoke-detector system installed

What is the probability that the smoke will be detected by only one device?

Expert Solution

Let A be the event that smoke is detected by device A and B be the event that smoke is detected by device B

P(A) = 0.95

P(B) = 0.98

P(A B) = 0.94

P(smoke will be detected by only one device) = P(smoke will be detected by A but not by B) + P(smoke will be detected by B but not by A)

= P(A B') + P(B A')

= [P(A) - P(AB)] + [P(B) - P(A B)]

= (0.95 - 0.94) + (0.98 - 0.94)

= 0.01 + 0.04

= 0.05

orchestra answered 2 years ago

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Question

A smoke-detector system uses two devices, A and B. If smoke is present, the probability that...

Solutions

Expert Solution

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