document.write( "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.
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\n" ); document.write( " Find the probability that the professor is chosen. \n" ); document.write( "
Algebra.Com's Answer #417271 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
here's the deal.
\n" ); document.write( "you have 11 staff members
\n" ); document.write( "there is 1 professor among the group.
\n" ); document.write( "the fact that 7 of the staff members are women is irrelevant to the problem therefore this is extraneous information that you can ignore.
\n" ); document.write( "i'll solve it 2 ways.
\n" ); document.write( "both ways will get the same answer which confirms that the answer has a high probability of being right.
\n" ); document.write( "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.
\n" ); document.write( "the formula for C(n,x) is n! divided by (x! * (n-x)!).
\n" ); document.write( "if you have a good scientific calculator that does combination formulas for you, you can also use that.\r
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\n" ); document.write( "\n" ); document.write( "anyway here goes with method 1.
\n" ); document.write( "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.
\n" ); document.write( "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.....
\n" ); document.write( "1 minus .45454545..... is equal to .54545454.....
\n" ); document.write( "that's the probability using method 1.\r
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\n" ); document.write( "\n" ); document.write( "method 2 is done as follows:
\n" ); document.write( "this method uses the combination formula.
\n" ); document.write( "the solution is the number of ways you can get the professor divided by the number of ways you can get anybody.
\n" ); document.write( "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.
\n" ); document.write( "C(11,6) is equal to 11! / (6! * 5!) which is equal to 462.
\n" ); document.write( "you need one professor and you have 1 to choose from so the number of ways you can get the professor is 1C1 = 1.
\n" ); document.write( "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.
\n" ); document.write( "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
\n" ); document.write( "the number of ways of getting the professor divided by the number of ways of getting anybody is therefore equal to 252/462 = .54545454....\r
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\n" ); document.write( "\n" ); document.write( "obviously method 1 was easier in this problem.
\n" ); document.write( "that may not be as true in other problems where you can't just get 1 minus the probability of not getting the professor.
\n" ); document.write( "both methods match so either method will get you the solution.
\n" ); document.write( "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.\r
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