Question

A particular automatic sprinkler system has two different types of activation devices for each 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.

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

b. What is the probability that the sprinkler head will not be activated?

c. What is the probability that both activation devices will work properly?

d. What is the probability that only the device with reliability 0.9 will work properly?

Answer 1

a) Therefore, the probability is P=0.98.

b) Therefore, the probability is P=0.02.

c) Therefore, the probability is P=0.72.

d) Therefore, the probability is P=0.18.

Explanation:

We know that: 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.

We get

`P(X)=0.9\\\\P(Y)=0.8\\`

a) We calculate the probability that the sprinkler head will be activated.

`P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.9+0.8-P(X)P(Y)\\\\P(X\cup Y)=1.7-0.9\cdot 0.8\\\\P(X\cup Y)=0.98\\`

Therefore, the probability is P=0.98.

b) We calculate the probability that the sprinkler head will not be activated.

`P=1-P(X\cup Y)\\\\P=1-0.98\\\\P=0.02\\`

Therefore, the probability is P=0.02.

c) We calculate the probability that both activation devices will work properly.

`P=P(X)\cdot P(Y)\\\\P=0.9\cdot 0.8\\\\P=0.72`

Therefore, the probability is P=0.72.

d) We calculate the probability that only the device with reliability 0.9 will work properly.

`P=P(X)\cdot P(Y^c)\\\\P=0.9\cdot 0.2\\\\P=0.18\\`

Therefore, the probability is P=0.18.