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if the balls are numbered from 1-100. and three balls are to be taken out by replacement then what is the probability that their sum will be odd.

A. 1/4
B. 3/4
C. 1/2
D. 5/8
E. 6/8




Since the sum is odd it should be (odd+odd+odd) or (odd+even+even). The probability to choose odd is always 1/2 since it is taken out with replacement. So the probability to choose (odd+odd+odd) is 1/2 * 1/2 * 1/2 = 1/8. Similarly the probability to choose (odd+even+even) is 1/2 * 1/2 * 1/2 = 1/8. But it could be (even+odd+even) or (even+even+odd). So the probability that their sum would be odd is 1/8 + 1/8 *3 = 4/8 = 1/2. The answer is, therefore, C

Another approach.
Since the chance to choose even or odd is equal(there are 50 odd balls and 50 even balls), the probability that their sum would be odd and the probability that their sum would be even are the same, moreover they are complementary events. So it should be 1/2 without tedious calculations.




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