Question

A box contains \(100\) balls, numbered from \(1\) to \(100\). If \(3\) balls are selected one after the other at random with replacement from the box, then the probability that the sum of the three numbers on the balls selected from the box is an odd number, is

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{3}{4}\)
• C. \(\dfrac{3}{8}\)
• D. \(\dfrac{1}{8}\)

Hint:

For the sum of numbers to be odd, the count of odd numbers selected must be odd.

Answer 1

The Correct Option is A

Step 1: Count odd and even numbered balls.
The balls are numbered from \(1\) to \(100\).
There are \(50\) odd numbers and \(50\) even numbers.
Therefore, for each draw,
\[ P(\text{odd})=\frac{50}{100}=\frac12 \] and
\[ P(\text{even})=\frac{50}{100}=\frac12 \]

Step 2: Understand when the sum is odd.
The sum of three selected numbers is odd when the number of odd numbers selected is odd.
So, possible cases are:
\[ 1 \text{ odd and } 2 \text{ even} \] or
\[ 3 \text{ odd} \]

Step 3: Calculate probability for exactly one odd number.
The probability of exactly one odd number in three draws is
\[ {}^3C_1\left(\frac12\right)^1\left(\frac12\right)^2 \]
\[ =3\left(\frac12\right)^3 \]
\[ =\frac38 \]

Step 4: Calculate probability for exactly three odd numbers.
The probability of three odd numbers is
\[ \left(\frac12\right)^3 \]
\[ =\frac18 \]

Step 5: Add the probabilities.
Required probability is
\[ \frac38+\frac18 \]
\[ =\frac48 \]
\[ =\frac12 \]

Step 6: Final conclusion.
Hence, the probability that the sum is odd is
\[ \boxed{\frac12} \]