document.write( "Question 1103356: 6 freshmen, 8 sophomores, 10 juniors, and 9 seniors are eligible to be on a committee. If a committee of 14 students is chosen at random, what is the probability that it is made up of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors? \r
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document.write( "Even if you don't use the exact problem, it's fine, since I want to learn how to find the probability. \n" );
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Algebra.Com's Answer #718043 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Okay; I will use different numbers: 4 freshmen, 3 sophomores, 7 juniors, and 7 seniors. And we will still make a committee of 14 with 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors. \n" ); document.write( "The basic definition of probability says that the answer is \n" ); document.write( "(number of ways of choosing 2 of the 4 freshmen AND 3 of the 3 sophomores AND 4 of the 7 juniors AND 5 of the 7 seniors) divided by (the total number of ways of choosing 14 of the total 21 students). \n" ); document.write( "Because the calculations involve choosing among candidates, you will be using the \"n choose r\" concept repeatedly. Since you are asking this question, I will assume you are familiar with that concept. \n" ); document.write( "The denominator of our probability fraction is the number of ways of choosing 14 of the 21 students: C(21,14). \n" ); document.write( "For the numerator, we have to choose certain numbers of each of the four class levels. So the numerator of the probability fraction will be C(4,2)*C(3,3)*C(7,4)*C(7,5). \n" ); document.write( "C(21,14) = 116280; C(4,2)*C(3,3)*C(7,4)*C(7,5) = 6*1*35*21 = 4410; the probability that the committee has 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors is \n" ); document.write( "Obviously the process for solving your example will be the same, just with different numbers. \n" ); document.write( " |