# Probability of stopping at a red light on way to school

On the way to school, Hendrik passes three traffic lights. For years he noted how often each light was red. He came to the following conclusions: light A: 50% chance of red light B: 30% chance of red light C: 10% chance of red Determine the probability distribution that shows the number of times that Hendrik has to stop on the way to school.

For now I found that P(x=0)=0.315 ($$P(\overline{A})$$ * $$P(\overline{B})$$ * $$P(\overline{C})$$) and P(x=3)=0.015 ($$P(A)$$ * $$P(B)$$ * $$P(C)$$). But I can't find P(x=1) and P(x=2)

Edit: These are the answers in my book $$P(x=0)=0.315; P(x=1)=0.485; P(x=2)=0.185; P(x=3)=0.015;$$

• Assuming these are all independent, P(1), the probability that one of the three lights is red, the other two are not, is Commented Jun 19, 2021 at 15:03

There is an interesting way to arrange the computation.

Let $$f(x) = (.5 + .5 x) (.7 + .3 x) (.9 + .1 x)$$.

If we expand $$f(x)$$, we get $$f(x) = 0.315\, +0.485 x+0.185 x^2+0.015 x^3$$

Notice a resemblance to your problem?

• This is pretty cool! But I don't really get why this is happening? Why would you turn it into a function and how does it give me the correct answer? Commented Jun 20, 2021 at 19:34
• @tomcajot If you carry out some of the computations by hand, as in user247327's answer, and also expand the polynomial by hand, you will see that the computations are exactly the same. So what we have done is not more efficient (It's just as much work), but it's a nice way to organize the computation. Unless you use a computer algebra system or Wolfram Alpha to do the expansion, and then the computer does the work. $f(x)$ is an example of a generating function. Commented Jun 21, 2021 at 12:39
• If you are interested in learning more about generating functions (they're really neat, IMHO), there are many resources here: math.stackexchange.com/questions/3142386/… Commented Jun 21, 2021 at 12:40
• Thanks a lot, I will be looking into it! Commented Jun 22, 2021 at 8:47

Assuming these are all independent, P(1), the probability that one of the three lights is red, the other two are not, is (1) the first is red and the other two are green: 0.5(0.7)(0.9)= 0.315. (2) the first is green, the second is red, the third is green: (0.5)(0.3)(0.9)= 0.135. (3) the first is green, the second is green, the third is red: (0.5)(0.7)(0.1)= 0.035. The probability exactly one light is red and the other two are green is 0.315+ 0.135+ 0.O35= 0.585.

The P(2), the probability that two lights are green and one is red is done the same way but reversing "red" and "green".