Question

Suppose that an employee at a local company checks his watch and realizes that he has 10 minutes to get to work on time. If he leaves now and does not get stopped by any traffic lights, he will arrive at work in exactly 8 minutes. In between his house and his work there are three traffic lights, A, B, and C. Each light that stops him will cause him to arrive an additional 2 minutes later. The following table displays the probability that he is stopped by each of the three traffic lights. Assume that the probability that he is stopped by any given light is independent of the probability that he is stopped by any other light.

What is the probability that the employee is not late for work?



Trafic light A P(A) 0.6
Trafic light B P(B) 0.4
Trafic light C P(C) 0.9


0.024 IS INCORRECT
Your answer represents the probability that the employee is stopped by zero lights and arrives at work in 8 minutes.
Although it is true that he will not be late for work if he is stopped by zero lights, he will also not be late for work if he is stopped by only one light.

Answer 1

The probability that the employee is not late to work is: 0.292

How to find the Probability?

If none of the traffic light stops him, he'll not be late to work.

So:

P(None) = P(A') * P(B') * P(C')

This gives:

P(None) = [1 - P(A)] * [1 - P(B)] * [1 - P(C)]

P(None) = [1 - 0.6] * [1 - 0.4] * [1 - 0.9]

P(None) = 0.024

If just one traffic light stops him, he'll not be late for work.

Because, he will get to work in 10 minutes

(i.e. 8 minutes + 2 minutes delay by 1 traffic light)

So:

P(One) = [P(A) * P(B') * P(C')] + [P(A') * P(B) * P(C')] + [P(A') * P(B') * P(C)]

P(One) = [0.6 * [1 - 0.4] * [1 - 0.9]] + [[1 - 0.6] * 0.4 * [1 - 0.9]] + [[1 - 0.6] * [1 - 0.4] * 0.9]

P(One) = 0.268

So, the probability that he is not late to work is:

P(not late) = 0.024 + 0.268

P(not late) = 0.292