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.)