document.write( "Question 1107060: A committee of size 5 is to be selected at random from 3 women and 5 men. Find the probability that committee consists of: (1)all men. (2)all women. \n" ); document.write( "
Algebra.Com's Answer #722060 by Edwin McCravy(20056)\"\" \"About 
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document.write( "(1)  First we find the numerator of the probability which is\r\n" );
document.write( "the number of way to choose 5 men \r\n" );
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document.write( "(5 men) CHOOSE (5 men) = 5C5 = 1 way\r\n" );
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document.write( "(That makes sense because we choose all the men in only 1 way.)\r\n" );
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document.write( "Now we find the denominator of the probability, which is the\r\n" );
document.write( "number of ways to choose 5 people from the 8 people\r\n" );
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document.write( "(8 people) CHOOSE (5 people) = 8C5 = 56 ways\r\n" );
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document.write( "So the probability is 1 way out of 56 ways or 1/56.\r\n" );
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document.write( "(2)  First we find the numerator of the probability which is\r\n" );
document.write( "the number of way to choose 5 women \r\n" );
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document.write( "(3 women) CHOOSE (5 women) = 3C5 = 0\r\n" );
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document.write( "(That makes sense because you can't choose 5 women because \r\n" );
document.write( "there are only 3 women.)\r\n" );
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document.write( "So the probability is 0 ways out of 56 ways or 0/56 or 0.\r\n" );
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document.write( "(This cannot happen because there aren't enough women to choose 5)\r\n" );
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document.write( "Edwin
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