This is (I believe) an incorrectly worded ripoff of an OG question. In the original question, I believe it is made clear that the strips are replaced into the bag after each drawing.
In this case, since there is a 5/10 = 1/2 chance of odd or even each time, and since there are as many ways to get an odd sum (OOO, OEE, EOE, EEO) as an even sum (EEE, EOO, OEO, OOE), there is a 1/2 chance that the sum is odd, and that it is even.
If the strips are NOT replaced, the math is a little different, but the answer is the same.
The probability of OOO is the same as EEE. Similarly P(OEE) = P(EOO), P(EOE) = P(OEO), and P(EEO) = P(OOE).
Since each odd sum has an equal match that is an even sum, the result is that each has a likelihood of 50% or 1/2.
You could go through and show that OOO = 5/10 * 4/9 * 3/8, but again, that is the same as EEE, so the totals in the "odd camp" will be the same as in the "even camp."
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