An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

Options

A. `1/32`
B. `5/16`
C. `3/16`
D. `1/2`

Answer

To solve the problem, we need to find the probability of getting an odd number an odd number of times when rolling a die \( n \) times, given that the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times.

### Step-by-Step Solution:

1. **Identify the probabilities of odd and even numbers:**
- A standard die has 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6).
- Therefore, the probability of rolling an odd number, \( P(O) \), is \( \frac{3}{6} = \frac{1}{2} \).
- Similarly, the probability of rolling an even number, \( P(E) \), is also \( \frac{3}{6} = \frac{1}{2} \).

2. **Set up the equation based on the given condition:**
- The probability of getting an odd number exactly 2 times when rolling the die \( n \) times is given by:
\[ P(X = 2) = \binom{n}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} = \binom{n}{2} \left(\frac{1}{2}\right)^n \]
- The probability of getting an even number exactly 3 times is given by:
\[ P(Y = 3) = \binom{n}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} = \binom{n}{3} \left(\frac{1}{2}\right)^n \]

3. **Set the two probabilities equal:**
\[ \binom{n}{2} \left(\frac{1}{2}\right)^n = \binom{n}{3} \left(\frac{1}{2}\right)^n \]
- Since \( \left(\frac{1}{2}\right)^n \) is common on both sides, we can cancel it out:
\[ \binom{n}{2} = \binom{n}{3} \]

4. **Use the property of combinations:**
- The equality \( \binom{n}{2} = \binom{n}{3} \) implies:
\[ n(n-1) = n(n-1)(n-2)/6 \]
- This simplifies to:
\[ 6 = n - 2 \quad \Rightarrow \quad n = 5 \]

5. **Find the probability of getting odd numbers an odd number of times:**
- We need to calculate the probability of getting odd numbers 1, 3, or 5 times when rolling the die 5 times:
\[ P(\text{odd times}) = P(X = 1) + P(X = 3) + P(X = 5) \]
- Calculate each term:
- For \( X = 1 \):
\[ P(X = 1) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^4 = 5 \cdot \frac{1}{2^5} = \frac{5}{32} \]
- For \( X = 3 \):
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{2^5} = \frac{10}{32} \]
- For \( X = 5 \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

6. **Combine the probabilities:**
\[ P(\text{odd times}) = \frac{5}{32} + \frac{10}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]

### Final Answer:
The probability of getting an odd number an odd number of times is \( \frac{1}{2} \).