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1-100 numbered balls... Probability

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by BarryLi » Wed Feb 23, 2011 12:49 pm
The question notes that this is with replacement, so that the probability of any ball being an even number or odd number is 1/2. This can be checked by noting whether the first number is an odd/even, and making sure the last number is the opposite. 1 is odd, 100 is even.

Let O and E represent an odd number and even number, respectively.
E+E+E = E
E+E+O = E+O+E = O+E+E = O
O+O+E = O+E+O = E+O+O = E
O+O+O = O

The probability of an odd sum is 4/8 = 1/2

OA: C
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by Night reader » Wed Feb 23, 2011 2:03 pm
The probability that we select three Balls which numbers sum to Odd number fall into four scenarios
1) Even+Even+Odd
2) Even+Odd+Even
3) Odd+Even+Even
4) Odd+Odd+Odd

The total possibilities for selecting Odd or Even --> 2*2*2 {fill in 3 slots/draws with 2 possible choices}, will result in total 8 possibilities and 1/8th chance of selecting Odd-Even with 3 draws of Balls

All four scenarios have equal probability of occurrence 1/8. Our required scenarios combined make 4* 1/8 = 1/2
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