Question

A computer software company receives hundreds of support calls each day. There are several common installation problems, call them A, B, C, and D. Several of these problems result in the same symptom, lock up after intiation. Suppose that the probability of a caller reporting the symptom lock up is 09 and the probability of a cater having problem A and a lock up is 03 Given that the caller reports a lock up, what is the probability that the cause is not problem A?

Answer 1

To find the probability that the cause is not problem A given that the caller reports a lock up, we can use conditional probability. The probability is approximately 0.7778 or 77.78%.

Step-by-step explanation:

To find the probability that the cause is not problem A, we can use the concept of conditional probability. Let's denote the event of the caller having problem A as event A and the event of the caller reporting a lock up as event B. We are given:

P(B) = 0.9 (probability of reporting a lock up)

P(A and B) = 0.3 (probability of having problem A and lock up)

We want to find P(not A | B), which represents the probability that the cause is not problem A given that the caller reports a lock up.

Using the formula for conditional probability: P(not A | B) = P(not A and B) / P(B)

Since not A and B are mutually exclusive events, P(not A and B) = 1 - P(A and B).

Substituting the values, we have:

P(not A | B) = (1 - P(A and B)) / P(B)

= (1 - 0.3) / 0.9

= 0.7 / 0.9

= 0.7778

Therefore, the probability that the cause is not problem A given that the caller reports a lock up is approximately 0.7778 or 77.78%.