This problem involves analyzing probabilities in a sequence of independent trials, which is a classic application of the Binomial Probability Distribution. We will first determine the total number of dice rolls ($n$) using the given condition and then calculate the required probability.

1. Key Concepts and Formulas

• Binomial Probability Distribution: Used for calculating the probability of exactly $k$ successes in $n$ independent Bernoulli trials. The formula is: $P(X=k) = {^n C_k} p^k (1-p)^{n-k}$ where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success in a single trial.
• Probability for a Fair Die: For an ordinary (fair) six-sided die, the probability of rolling an odd number (1, 3, 5) is $p_{\text{odd}} = 3/6 = 1/2$. The probability of rolling an even number (2, 4, 6) is $p_{\text{even}} = 3/6 = 1/2$.
• Binomial Coefficient Property: If ${^n C_k} = {^n C_m}$ and $k \neq m$, then $k+m=n$. This property is crucial for solving for $n$.
• Sum of Odd/Even Terms in Binomial Expansion: For a binomial expansion $(q+p)^n$, the sum of probabilities for an odd number of successes is given by $\frac{1}{2}[(q+p)^n - (q-p)^n]$. When $q=1-p$, this simplifies to $\frac{1}{2}[1 - (1-2p)^n]$.

2. Step-by-Step Solution

Step 1: Determine the Total Number of Rolls ($n$)

Let $n$ be the total number of times the dice is rolled. For an ordinary dice, the probability of getting an odd number is $p = 1/2$, and the probability of getting an even number is $q = 1 - p = 1/2$.

The problem states that "the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times."

1. Probability of getting an odd number 2 times: Here, 'success' is getting an odd number ($p=1/2$). We want $k=2$ successes. Using the binomial formula: $P(\text{odd 2 times}) = {^n C_2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} = {^n C_2} \left(\frac{1}{2}\right)^{n}$

2. Probability of getting an even number 3 times: Here, 'success' is getting an even number ($p=1/2$). We want $k=3$ successes. Using the binomial formula: $P(\text{even 3 times}) = {^n C_3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} = {^n C_3} \left(\frac{1}{2}\right)^{n}$

3. Equate the probabilities and solve for $n$: Given that these probabilities are equal: ${^n C_2} \left(\frac{1}{2}\right)^n = {^n C_3} \left(\frac{1}{2}\right)^n$ Since $\left(\frac{1}{2}\right)^n \neq 0$, we can cancel it from both sides: ${^n C_2} = {^n C_3}$ Using the binomial coefficient property ${^n C_k} = {^n C_m} \Rightarrow k=m \text{ or } k+m=n$. Since $2 \neq 3$, we must have $2+3=n$. Therefore, $n=5$. The dice is rolled 5 times.

Step 2: Calculate the Probability of Getting an Odd Number for an Odd Number of Times

Now we know $n=5$ and the probability of getting an odd number in a single roll is $p=1/2$. We need to find the probability that an odd number appears an odd number of times. This means we need to sum the probabilities for $k=1$, $k=3$, and $k=5$ successes.

Using the general formula for the sum of odd terms in a binomial expansion: $P(\text{odd occurrences}) = \frac{1}{2}[1 - (1-2p)^n]$ Substitute $n=5$ and $p=1/2$: $P(\text{odd occurrences}) = \frac{1}{2}\left[1 - \left(1-2\left(\frac{1}{2}\right)\right)^5\right]$ $P(\text{odd occurrences}) = \frac{1}{2}\left[1 - (1-1)^5\right]$ $P(\text{odd occurrences}) = \frac{1}{2}\left[1 - (0)^5\right]$ $P(\text{odd occurrences}) = \frac{1}{2}[1 - 0]$ $P(\text{odd occurrences}) = \frac{1}{2}$

Alternatively, calculate each term:

• $P(X=1) = {^5 C_1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^4 = 5 \times \frac{1}{32} = \frac{5}{32}$
• $P(X=3) = {^5 C_3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \times \frac{1}{32} = \frac{10}{32}$
• $P(X=5) = {^5 C_5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0 = 1 \times \frac{1}{32} = \frac{1}{32}$

Summing these probabilities: $P(\text{odd occurrences}) = \frac{5}{32} + \frac{10}{32} + \frac{1}{32} = \frac{5+10+1}{32} = \frac{16}{32} = \frac{1}{2}$

3. Common Mistakes & Tips

• Misinterpreting "Ordinary Dice": An "ordinary dice" implies a fair die where each face has a probability of $1/6$. Consequently, the probability of an odd number (1, 3, 5) is $3/6 = 1/2$, and an even number (2, 4, 6) is also $3/6 = 1/2$. Do not assume a loaded die unless specified.
• Incorrectly Solving for $n$: Always remember the property ${^n C_k} = {^n C_m} \Rightarrow k=m \text{ or } k+m=n$. Expanding factorials can be time-consuming and prone to errors.
• Missing the Shortcut for $p=1/2$: When $p=1/2$, the binomial distribution is symmetric. For any $n$, the sum of probabilities for an odd number of successes is equal to the sum of probabilities for an even number of successes. Since the total probability is 1, each sum must be $1/2$. This is a powerful shortcut that directly leads to the answer $1/2$ once $n$ is determined to be 5 (or any integer for that matter).

4. Summary

We identified the problem as an application of the Binomial Probability Distribution. By setting up equations based on the given conditions and using the properties of binomial coefficients and fair dice, we determined that the total number of rolls is $n=5$. Since the probability of getting an odd number on a fair die is $p=1/2$, the distribution is symmetric. For a symmetric binomial distribution, the probability of getting an odd number of successes (odd occurrences) is always $1/2$.

5. Final Answer

The final answer is $\boxed{\text{C}}$.