A box contains 100 balls, numbered from 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?

Solution(By Examveda Team)

$$P({\text{odd}}) = P({\text{even}}) = \frac{1}{2}$$     (because there are 50 odd and 50 even numbers)
Sum or the three numbers can be odd only under the following 4 scenarios:
$$\eqalign{ & {\text{odd}} + {\text{odd}} + {\text{odd}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{odd}} + {\text{even}} + {\text{even}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{even}} + {\text{odd}} + {\text{even}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{even}} + {\text{even}} + {\text{odd}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr} $$
Other combinations of odd and even will give even numbers.
Adding up the 4 scenarios above:
$$\eqalign{ & = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \cr & = \frac{4}{8} \cr & = \frac{1}{2} \cr} $$