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
Step 1 of 4
Let A be the event that the first type with reliability is 0.9
Let B be the event that the second type with reliability is 0.8
a) P(sprinkler head will activate)=P(AUB)
=P(A)+P(B)-P(AB)
=P(A)+P(B)-P(A) P(B) (Both are independent)
=0.9+0.8-(0.9)(0.8)
=0.98
