SOLUTION: You have a traffic light of only a red and a green light. What is the probability that you will get 2 green lights out of 4 times the light changes?
I figured that probability of
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Question 222179: You have a traffic light of only a red and a green light. What is the probability that you will get 2 green lights out of 4 times the light changes?
I figured that probability of one try is .5. It is either green or red the 1st time. Then I wrote out the possibilities of 2 light changes, and came up with rr, rg, gr, gg. (So 2 greens is 1/4, or .5 x .5 = .25.) After this, I am stuck. I don't think it's .5 x .5 x .5 x .5. That would be too small a probability. I think the events are independent and mutually exclusive, but then I don't know what to do. My dad and I tried to write out all the possibilities, and got 16 of them, with 6 times getting 2 greens, so that would be 6/16 or 3/8, but we're not sure. We did not count the cases where we got 3 or 4 greens... do you count those too? And I don't have a good formula to go by... if the problem were 9 tries or something, I'd never figure it out!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
You have a traffic light of only a red and a green light. What is the probability that you will get 2 green lights out of 4 times the light changes?
I figured that probability of one try is .5. It is either green or red the 1st time. Then I wrote out the possibilities of 2 light changes, and came up with rr, rg, gr, gg. (So 2 greens is 1/4, or .5 x .5 = .25.) After this, I am stuck. I don't think it's .5 x .5 x .5 x .5. That would be too small a probability. I think the events are independent and mutually exclusive, but then I don't know what to do. My dad and I tried to write out all the possibilities, and got 16 of them, with 6 times getting 2 greens, so that would be 6/16 or 3/8, but we're not sure. We did not count the cases where we got 3 or 4 greens... do you count those too? And I don't have a good formula to go by... if the problem were 9 tries or something, I'd never figure it out!
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I think you and your Dad did well.
It's a binomial problem with n = 4, p = 1/2 , x = 2
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P(x=2) = 4C2*(1/2)^2*(1/2)^2
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P(x=2) = 6/16 = 3/8
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It's the same as having exactly 2 girls born in a family of 4.
The events (red and green) are independent because you don't know
whether you will get a red or a green at each intersection.
Similar to tossing a coin.
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Cheers,
Stan H.
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