Question 671115: An exam paper has 6 questions. Each question will be marked by a different person. The markers will be randomly chosen from 11 academic staff, of whom 7 are women. One of the staff members is a professor.
Find the probability that the professor is chosen.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! here's the deal.
you have 11 staff members
there is 1 professor among the group.
the fact that 7 of the staff members are women is irrelevant to the problem therefore this is extraneous information that you can ignore.
i'll solve it 2 ways.
both ways will get the same answer which confirms that the answer has a high probability of being right.
combination formula used is C(n,x) where n is the total possible choices you have and x is the total possible choices that you want.
the formula for C(n,x) is n! divided by (x! * (n-x)!).
if you have a good scientific calculator that does combination formulas for you, you can also use that.
anyway here goes with method 1.
the probability that you will get the professor for one of the 6 questions is equal to 1 minus the probability that you will not get the professor for one of the 6 questions.
that probability is equal to 10/11 * 9/10 * 8/9 * 7/8 * 6/7 * 5/6 which is equal to (10! / 4!) / (11! / 5!) which is equal to .45454545.....
1 minus .45454545..... is equal to .54545454.....
that's the probability using method 1.
method 2 is done as follows:
this method uses the combination formula.
the solution is the number of ways you can get the professor divided by the number of ways you can get anybody.
the number of ways you can get anybody is equal to C(11,6) which is the number of ways you can get any of the 11 staff members to be on a team of 6.
C(11,6) is equal to 11! / (6! * 5!) which is equal to 462.
you need one professor and you have 1 to choose from so the number of ways you can get the professor is 1C1 = 1.
you need 5 other members on the team of 6 that are not professors and you have 10 to choose from so the number of ways you can get the additional 5 members that are not professors is C(10,5) = 252.
the total number of ways you can get the professor and the non-professors to be on the team of 6 is therefore equal to 1 * 252 = 252
the number of ways of getting the professor divided by the number of ways of getting anybody is therefore equal to 252/462 = .54545454....
obviously method 1 was easier in this problem.
that may not be as true in other problems where you can't just get 1 minus the probability of not getting the professor.
both methods match so either method will get you the solution.
the second method is trickier but instructive because sometime you are taught to do it this way and it helps to understand how to do it.
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