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**Boris B** I have a solution to a practice problem that took me a long time, and doesn't really match any of my multiple choices.

**50% of drivers are low-risk, 30% are moderate risk, and 20% are high risk. For any given group of 4 drivers, what are the odds that the number of high risk drivers is 2 greater than the number of low?** (I called this outcome "success".)

**What is the fastest way to do this?** My ways seemed tedious and make me suspect I'm missing something.

At first I thought I should use means and standard deviations and overlay bell curves, but then I though that was going to give me a lot of fractional drivers which don't make any sense.

Then I decided I'd use the cumulative distribution function. First, I calculated the odds of their being each of the five possibilities for number of high-risk drivers. I found:

PH(0) =0.410 (failure)

PH(1) =0.410 (failure)

PH(2) =0.154 (see below)

PH(3) =0.026 (success)

PH(4) =0.002 (success)

After each number, I noted whether that outcome indicated fulfilled the condition.

For the 2 high-risk scenario, I calculated a 0.625 chance that each of the remaining drivers would be low risk. (This is intuitive based on the 3 to 5 ratio of non-high-risk drivers, but I'm not sure I have the math proof down.)

(I know the below notation is probably not right, but maybe you can make sense of it.)

P(L=2|H=2) =0.391 (failure)

P(M=2|H=2) =0.141 (success)

P(L=1 and M=1|H=2) =0.468 (failure)

I added the probabilities of my three success conditions and found:

0.026 + 0.002 + (0.154x0.141)= 0.050

I'm doubting this is the correct answer, because it is only moderately close to one of the choices provided (0.06). (Let me know if I need to show more work; I am pretty sure I got the binomial probability distribution function correct, but I may have gotten mixed up in one of the exponents.)