[Math] Number of ways to form committee of 5 people

combinationsprobability

I am trying to solve the following problem on combinations:

You wish to select five persons from seven men and six women to form a
committee that includes at least three men.

a. In how many ways can
you form the committee?

b. If you randomly choose five persons to form
the committee, what is the probability that you will get a committee
with at least three men?

My attempt:

a) Number of ways = $ C(7,3) * C(10,2) = 1575 $
b) Sample space = $ C(13,5) = 1287 $

Now the working for (b) looks pretty strange because the probability will be $ \frac{1575}{1287} > 1 $ which is definitely incorrect. Can anyone please advise me what is wrong here?

Best Answer

(a) is incorrect, because you count some selections of men twice. For example, picking the first five men is counted in:

  • I first pick the first three men, then the fourth and fifth
  • I first pick men 1,2,4, then 3,5
  • I first pick 1,2,5, then 3,4

and so on. So, basically, in your solution to (a), you made a distinction between "men chosen as one of the first three" and "men chosen as one of the final two", a distinction that doesn't really exist.

So your answer for (a) is not correct. I suggest just summing the numbers for a commitee with exactly three, four or five men.

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