Question 977814: I've run into a road block with Combinatorial Probability. I did the first two questions on my homework fine, I think the idea here is to add a step. Do not know what the next step is (meaning, the material is progressively more difficult). I computed previously leading to the n!/r!(n-r)!.
3. There are 27 tasks available. You will be randomly assigned 12 of them. There are 10
of them that you want to do.
(a) What is the probability that you will be assigned none of the tasks that you want?
(b) What is the probability that you will be assigned exactly 7 of the tasks that you
want?
(c) What is the probability that you will be assigned at least 1 of the tasks that you
want?
(d) What is the probability that you will be assigned at least 7 of the tasks that you
want?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Part a) There are 10 tasks that you want to do, so there are 17 tasks you don't want to do.
The probability of being assigned, for the first assignment, a task you don't want to do is then 
Then, based on the assumption that you actually received a "don't want to do" task for the first assignment, then you now have 16 tasks left that you don't want to do out of a total of 26 tasks, hence 
Then, 
Then, 
And so on for 17 factors. The combined probability is the product of all 17 factors.
John
My calculator said it, I believe it, that settles it
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