Question

A particular automatic sprinkler system has two different types of activation devices for each sprinkler head. One type has a reliability of 0.89; that is, the probability that it will activate the sprinkler when it should is 0.89. The other type, which operates independently of the first type, has a reliability of 0.79. If either device is triggered, the sprinkler will activate. Suppose a fire starts near a sprinkler head. What is the probability that the sprinkler head will be activated

Answer 1

0.70

Explanation:

Let the event A be "the first device is triggered" and the event B be "the second device is a". We can state each probability as:

P(A) = 0.89

P(B) = 0.79

We want to determine the probability that the sprinkler head is activated given any of the devices is triggered, that is, P(A ∩ B).

For independent events, we have:

P(A ∩ B) = P(A) × P(B)

P(A ∩ B) = 0.89 × 0.79 = 0.70