A dice is thrown `(2n+1)` times. The probability that faces with even numbers show odd number of times is

Options

A. `(2n + 1)/(2n + 3)`
B. less than `(1)/(2)`
C. Greater than `(1)/(2)`
D. None of these

Answer

To solve the problem, we need to find the probability that the faces with even numbers show an odd number of times when a die is thrown \(2n + 1\) times.

### Step-by-Step Solution:

1. **Identify the Even Faces on a Die:**
The even faces on a standard die are 2, 4, and 6. Therefore, there are 3 even outcomes.

2. **Calculate the Probability of Rolling an Even Number:**
The total number of outcomes when rolling a die is 6 (1, 2, 3, 4, 5, 6). The probability \(P\) of rolling an even number is:
\[ P(\text{even}) = \frac{\text{Number of even outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \]

3. **Calculate the Probability of Rolling an Odd Number:**
Since there are 3 odd outcomes (1, 3, 5), the probability \(Q\) of rolling an odd number is:
\[ Q(\text{odd}) = 1 - P(\text{even}) = 1 - \frac{1}{2} = \frac{1}{2} \]

4. **Define the Random Variable:**
Let \(X\) be the number of times an even number is rolled in \(2n + 1\) trials. \(X\) follows a binomial distribution with parameters \(n = 2n + 1\) and \(p = \frac{1}{2}\).

5. **Find the Probability of Odd Occurrences of Even Numbers:**
We need to calculate the probability that \(X\) is odd. This can be expressed as:
\[ P(X \text{ is odd}) = P(X = 1) + P(X = 3) + P(X = 5) + \ldots + P(X = 2n + 1) \]
Using the binomial probability formula:
\[ P(X = r) = \binom{2n + 1}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{(2n + 1 - r)} = \binom{2n + 1}{r} \left(\frac{1}{2}\right)^{2n + 1} \]

6. **Summing the Probabilities:**
The probability that \(X\) is odd can be simplified using the property of binomial coefficients:
\[ P(X \text{ is odd}) = \frac{1}{2^{2n + 1}} \sum_{r \text{ odd}} \binom{2n + 1}{r} \]
The sum of the binomial coefficients for odd \(r\) can be expressed as:
\[ \sum_{r \text{ odd}} \binom{2n + 1}{r} = 2^{2n} \]
Therefore:
\[ P(X \text{ is odd}) = \frac{1}{2^{2n + 1}} \cdot 2^{2n} = \frac{1}{2} \]

### Final Result:
Thus, the probability that faces with even numbers show an odd number of times is:
\[ \frac{1}{2} \]