An ordinary dice is rolled a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even numebr 3 times. Then the probability of getting an odd number and odd number of times, is

Options

A. `1//32`
B. `5//16`
C. `1//2`
D. none of these

Explanation

To solve the problem, we need to find the probability of getting an odd number an odd number of times when an ordinary die is rolled a certain number of 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. **Define the Variables**:
Let \( n \) be the number of times the die is rolled. The probability of getting an odd number when rolling a die is \( p = \frac{3}{6} = \frac{1}{2} \) and the probability of getting an even number is \( q = \frac{3}{6} = \frac{1}{2} \).

2. **Set Up the Probabilities**:
The probability of getting an odd number exactly 2 times is given by the binomial probability formula:
\[ P(\text{odd 2 times}) = \binom{n}{2} p^2 q^{n-2} \]
Substituting \( p \) and \( q \):
\[ P(\text{odd 2 times}) = \binom{n}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{n-2} = \binom{n}{2} \left(\frac{1}{2}\right)^n \]

Similarly, the probability of getting an even number exactly 3 times is:
\[ P(\text{even 3 times}) = \binom{n}{3} p^3 q^{n-3} \]
Substituting \( p \) and \( q \):
\[ P(\text{even 3 times}) = \binom{n}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{n-3} = \binom{n}{3} \left(\frac{1}{2}\right)^n \]

3. **Set the Probabilities Equal**:
Since the problem states that these two probabilities are equal, we can set them equal to each other:
\[ \binom{n}{2} \left(\frac{1}{2}\right)^n = \binom{n}{3} \left(\frac{1}{2}\right)^n \]
We can cancel \( \left(\frac{1}{2}\right)^n \) from both sides (as long as \( n \geq 3 \)):
\[ \binom{n}{2} = \binom{n}{3} \]

4. **Use the Binomial Coefficient Formula**:
The binomial coefficients can be expressed as:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
\[ \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Setting these equal gives:
\[ \frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6} \]

5. **Solve for \( n \)**:
Cross-multiplying to eliminate the fractions:
\[ 6n(n-1) = 2n(n-1)(n-2) \]
Dividing both sides by \( n(n-1) \) (assuming \( n \neq 0, 1 \)):
\[ 6 = 2(n-2) \]
Simplifying gives:
\[ 6 = 2n - 4 \implies 2n = 10 \implies n = 5 \]

6. **Calculate the Probability of Getting an Odd Number an Odd Number of Times**:
Now, we need to find the probability of getting an odd number of odd outcomes. The possible odd outcomes are 1, 3, or 5. We can calculate:
\[ P(\text{odd 1 time}) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{4} = \binom{5}{1} \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32} \]
\[ P(\text{odd 3 times}) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{2} = \binom{5}{3} \left(\frac{1}{2}\right)^5 = 10 \cdot \frac{1}{32} = \frac{10}{32} \]
\[ P(\text{odd 5 times}) = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

Now, summing these probabilities gives:
\[ P(\text{odd number of odd outcomes}) = P(\text{odd 1 time}) + P(\text{odd 3 times}) + P(\text{odd 5 times}) = \frac{5}{32} + \frac{10}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} \]

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