Hi,
There are 5 even and 5 odd in 1 to 10.
For the sum to be odd, either all 3 should be odd or 1 odd,2even.
all 3 odd -> 5C3
1 odd,2 even -> 5C1*5C2
Number of ways of selecting 3 numbers from 10 is 10C3
So, probability is (5C3 + 5C2*5C1)/10C3 = (10+50)/120 = 1/2

Hence, D

There're 5 odd numbers, so probability of 1 odd = 5/10 = 1/2

sum is odd can happen in 4 ways:

odd + odd +odd = (1/2)^3

even+even+odd

even+odd+even

odd+even+even

total probability = 4 * (1/2)^3 = 4/8 = 1/2

In case of favorable no. of cases we have

x1 + x2 + x3 = odd in only two cases.

all three are odd.
two are even and one is odd.

Therefore, we have

1. 2 even, 1 odd ----> 5c1*5c2
2. 3 odd --->>> 5c3

total no. of cases = 10c3

Hence prob. = (5c1*5c2 + 5c3) /10c3 = 1/2

so my answer is d

I am really confused here. Since there are 5 odd numbers:

all three odd: 5 x 4 x 3 = 60 desired outcomes
one odd, two even: 5 x 5 x 4 = 100 desired outcomes

Total desired outcomes = 160, no? Experts?

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."