Question

An apartment has two fire alarms. In the event of a fire, the probability that alarm A will fail is 0.004, and the probability that alarm B will fail is 0.01. Assume the two failures are independent, what is the probability that at least one alarm fails in the event of a fire

Answer 1

Answer: 0.0136

Explanation:

Let events are:

A = alarm A will fail

B= alarm B will fail

We have given ,

P(A) = 0.004 , P(B)=0.01

Since both events are independent , so

P(A and B) = P(A) x P(B)

= 0.04 x 0.01 =0.0004

i.e. P(A and B) =0.0004

Now , P(A or B) = P(A)+P(B)-P(A and B)

= 0.004+ 0.01-0.0004=0.0136

Hence, the probability that at least one alarm fails in the event of a fire is 0.0136 .