On her way to work, a commuter encounters four traffic signals. Assume that the distance between each of the four is sufficiently great that her probability of getting a green light at any intersection is independent of what happened at any previous intersection. The first two lights are green for forty seconds of each minute; the last two, for thirty seconds of each minute. What is the probability that the commuter has to stop at least three times?

7 in 36 or 0.1944

Step-by-step explanation:

The probability of having to stop at least three times is the probability of getting 3 or 4 red lights.

For the first two lights, the probability of getting them red is 20 in 60 (1/3).

For the last two lights, the probability of getting them red is 30 in 60 (1/2).

The probability of all of them being red is:

The probability of three of them being red (3 red + 1 green) is:

Therefore, the probability of at least three red lights is:

The probability is 7 in 36 or 0.1944.

2

step-by-step explanation:

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