Question

A particular automatic sprinkler system has two different types of activation devices for each 3 sprinkler head. One type has a reliability of 0.9; that is, the probability that it will activate the sprinkler when it should is 0.9. The other type, which operates independently of the first type, has a reliability of 0.8. 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

Answer:

There is an 85% probability that the sprinkler head will be activated.

Step-by-step explanation:

We have these following probabilities:

50% of the first sprinkler being triggered

90% of the first sprinkler being reliable

50% of the second sprinkler being triggered

80% of the second sprinkler being reliable.

What is the probability that the sprinkler head will be activated?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability of the first sprinkler being triggered and activated. So:

$P_{1} = 0.5*0.9 = 0.45$

$P_{2}$ is the probability of the second sprinkler being triggered and activated. So:

$P_{2} = 0.5*0.8 = 0.40$

$P = P_{1} + P_{2} = 0.40 + 0.45 = 0.85$

There is an 85% probability that the sprinkler head will be activated.