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.

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MrRoyal MrRoyal · Answer 1 · 2021-09-13T03:50:55+02:00

Probabilities are used to determine the possibility of an event. The probability that the employee is not late to work is 0.604.

The probability of each stop is given as:

[tex]P(A) =0.3[/tex]

[tex]P(B) =0.4[/tex]

[tex]P(C) =0.6[/tex]

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

So:

[tex]P(None) = P(A') \times P(B') \times P(C')[/tex]

This gives:

[tex]P(None) = [1 - P(A)] \times [1 - P(B)] \times [1 - P(C)][/tex]

[tex]P(None) = [1 - 0.3] \times [1 - 0.4] \times [1 - 0.6][/tex]

[tex]P(None) = 0.168[/tex]

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:

[tex]P(One) = [P(A) \times P(B') \times P(C')] +[P(A') \times P(B) \times P(C')] + [P(A') \times P(B') \times P(C)][/tex]

This gives:

[tex]P(One) = [0.3 \times (1 - 0.4) \times (1 - 0.6)] +[(1 - 0.3) \times 0.4 \times (1 - 0.6)] + [(1 - 0.3) \times (1 - 0.4) \times 0.6][/tex]

[tex]P(One) = [0.072] +[0.112] + [0.252][/tex]

[tex]P(One) = 0.436[/tex]

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

[tex]P(Not\ Late) = P(None) + P(One)[/tex]

[tex]P(Not\ Late) = 0.168 + 0.436[/tex]

[tex]P(Not\ Late) = 0.604[/tex]

Hence, the probability that the employee is not late to work is 0.604