A sprinkler system has two different independent types of ac

A sprinkler system has two different, independent types of activation devices for each sprinkler head. One type has a reliability of 0.9, while the other type has a reliability of 0.8. if either device is triggered , the sprinkler will activate. Suppose a fire starts near a sprinkler system:

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?

Solution

given

A sprinkler system has two different, independent types of activation devices for each sprinkler head. One type has a reliability of 0.9, while the other type has a reliability of 0.8. if either device is triggered , the sprinkler will activate.

let A be device with realibility 0.9

B be device with realibility 0.8

A) we have to find  probability that the sprinkler head will be activated

means we have to find P(A U B) since A and B are independent hence P(A intersection B)=P(A)*P(B)

since P(A U B)=P(A)+P(B)-P(A intersection B)

=0.9+0.8-(0.8*0.9)=1.7-0.72=0.98

B) we have to find probability that the sprinkler head will not be activated

means we have to find the P(A U B)^c

required probability=1-P(A U B)=1-0.98=0.02

C)we have to find  probability that both activation devices will work properly

means we have to find P(A intersection B)

hence required probability =P(A)*P(B)=0.9*0.8=0.72

D) we have to find probability that only the device with reliability 0.9 will work properly

means we have to find P(A intersection B^c )

since P(B^c)=1-P(B)=1-0.8=0.2

since A and B are independent hence A and B^c is also dependent

hence P(A intersection B^c)=P(A)*P(B^c)=0.9*0.2=0.18 answer