document.write( "Question 356933: A box contains eleven balls, numbered 1,2,3….10,11,11. if 6 balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd? \n" ); document.write( "

Algebra.Com's Answer #254897 by robertb(5830) $\"\"$ $\"About$

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Breaking down the possibilities:
\n" ); document.write( "0 odd, 6 even = sum is even
\n" ); document.write( "1 odd, 5 even = sum is odd
\n" ); document.write( "2 odd, 4 even = sum is even
\n" ); document.write( "3 odd, 3 even = sum is odd
\n" ); document.write( "4 odd, 2 even = sum is even
\n" ); document.write( "5 odd, 1 even = sum is odd
\n" ); document.write( "6 odd, 0 even = sum is even
\n" ); document.write( "We only need to consider the odd cases given above
\n" ); document.write( "i) 1 odd, 5 even. The number of ways of selecting 1 odd and 5 even is
\n" ); document.write( " $\"C%286%2C1%29%2AC%285%2C5%29+=+6\"$
\n" ); document.write( "ii)3 odd, 3 even. The number of ways of selecting 3 odd and 3 even is
\n" ); document.write( " $\"C%286%2C3%29%2AC%285%2C3%29+=+200\"$
\n" ); document.write( "iii)5 odd, 1 even. The number of ways of selecting 5 odd and 1 even is
\n" ); document.write( " $\"C%286%2C5%29%2AC%285%2C1%29+=+30\"$ .
\n" ); document.write( "Now the total number of ways of selecting 6 out of 11 balls is $\"C%2811%2C6%29+=+462\"$ . Therefore the probability that the sum of the numbers on the balls drawn is odd is $\"%286%2B200%2B30%29%2F462+=+236%2F462+=+118%2F231\"$ . (Since the 3 cases above are mutually exclusive.)\r
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