Problem Solving

Can you help me understand and solve below problem? Thanks

A box contains 100 balls numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?

a)3/8
b)1/2
c)5/8
d)1/4
e)3/4

Approach 1:

The sum will be odd if all 3 numbers selected are odd or if exactly 1 number selected is odd.
Let O = odd and E = even.
Since there are an equal number of odd numbers and even numbers in the box, P(O) = 1/2 and P(E) = 1/2.

Probability all 3 are odd:
P(OOO) = 1/2 * 1/2 * 1/2 = 1/8

Probability exactly 1 is odd:
P(OEE) = 1/2 * 1/2 * 1/2 = 1/8
Since EOE and EEO are also good outcomes, we multiply this result by 3:
3 * 1/8 = 3/8

Since we'll get a good outcome if all 3 numbers selected are odd OR if exactly 1 number selected is odd, we add the fractions:
1/8 + 3/8 = 4/8 = 1/2.

The correct answer is B.

Approach 2:

Since there are 50 odd numbers and 50 even numbers:
P(odd number) = 1/2.
P(even number) = 1/2.

Ways to get an ODD sum:
3 odd numbers
1 odd number, 2 even numbers

Ways to get an EVEN sum:
3 even numbers
1 even number, 2 odd numbers

Since P(odd) = P(even) = 1/2:
P(3 odds) = P(3 evens).
P(1 odd, 2 evens) = P(1 even, 2 odds).

Adding the two equations, we get;
P(3 odds) + P(1 odd, 2 evens) = P(3 evens) + P(1 even, 2 odds)
P(odd sum) = P(even sum).

Since the two probabilities are equal, and the sum must be odd or even:
P(odd sum) = 1/2.
P(even sum) = 1/2.

The correct answer is B.