document.write( "Question 1157466: A committee of three is chosen at random from a group of four women and five men. Find the probability that the committee contains at least one man. \n" ); document.write( "
Algebra.Com's Answer #780334 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "P(at least one man) = 1 - P(no men) = 1-P(3 women) \n" ); document.write( "P(3 women and no men) = (C(4,3)*C(5,0))/C(9,3) = (4*1).84 = 4/84 = 1/21 \n" ); document.write( "P(at least one man) = 1-1/21 = 20/21 \n" ); document.write( "You can get practice calculating probabilities like this by finding the probabilities of choosing 1, 2, or 3 men. \n" ); document.write( "P(1 man) = (C(4,2)*C(5,1))/C(9,3) = (6*5)/84 = 30/84 \n" ); document.write( "P(2 men) = (C(4,1)*C(5,2))/C(9,3) = (4*10)/84 = 40/84 \n" ); document.write( "P(3 men) = (C(4,0)*C(5,3))/C(9,3) = (1*10)/84 = 10/84 \n" ); document.write( "The sum of the probabilities is (4+30+40+10)/84 = 1. \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |