Answer:
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:
[tex]P(R=4) = \frac{1}{3}*\frac{1}{3}*\frac{1}{2}*\frac{1}{2}\\P(R=4) =\frac{1}{36}=\frac{1}{36}[/tex]
The probability of three of them being red (3 red + 1 green) is:
[tex]P(R=3) = {(1-\frac{1}{3})*\frac{1}{3}*\frac{1}{2}}*\frac{1}{2}+(1-\frac{1}{3})*\frac{1}{3}*\frac{1}{2}*\frac{1}{2}+\frac{1}{3}*\frac{1}{3}*(1-\frac{1}{2})*\frac{1}{2}+\frac{1}{3}*\frac{1}{3}*(1-\frac{1}{2})*\frac{1}{2}\\P(R=3) =2*(\frac{2}{3}*\frac{1}{3}*\frac{1}{2}*\frac{1}{2})+2*(\frac{1}{3}*\frac{1}{3}*\frac{1}{2}*\frac{1}{2})\\ P(R=3) =\frac{4}{36}+\frac{2}{36}\\ P(R=3) =\frac{6}{36}[/tex]
Therefore, the probability of at least three red lights is:
[tex]P=\frac{1}{36}+\frac{6}{36}=\frac{7}{36}\\ P=0.1944[/tex]
The probability is 7 in 36 or 0.1944.
The probability that the commuter has to stop at least three times is at least 5.55%.
Since on her way to work, a commuter encounters four traffic signals, assuming 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, and the first two lights are green for forty seconds of each minute; the last two, for thirty seconds of each minute, to determine what is the probability that the commuter has to stop at least three times the following calculation must be performed:
Therefore, the probability that the commuter has to stop at least three times is at least 5.55%.
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