A fair coin is tossed 99 times. If X is the number of times heads occur, then `P(X=r)` is maximum when `r` is `49 , 50` b. `50 , 51` c. `51 , 52` d. none of these

Options

A. 49
B. 52
C. 51
D. 50

Explanation

To solve the problem, we need to determine the values of \( r \) for which the probability \( P(X = r) \) is maximized when a fair coin is tossed 99 times.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a fair coin 99 times, and we want to find the values of \( r \) (the number of heads) that maximize the probability \( P(X = r) \).

2. **Using Binomial Distribution**:
The number of heads in 99 tosses follows a binomial distribution, where:
- \( n = 99 \) (the number of trials),
- \( p = \frac{1}{2} \) (the probability of getting heads),
- \( q = \frac{1}{2} \) (the probability of getting tails).

The probability mass function for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
Substituting the values, we have:
\[ P(X = r) = \binom{99}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{99-r} = \binom{99}{r} \left(\frac{1}{2}\right)^{99} \]

3. **Maximizing the Probability**:
To maximize \( P(X = r) \), we need to maximize \( \binom{99}{r} \). The binomial coefficient \( \binom{n}{r} \) is maximized when \( r \) is around \( \frac{n}{2} \).

4. **Calculating \( \frac{n}{2} \)**:
For \( n = 99 \):
\[ \frac{99}{2} = 49.5 \]
Since \( r \) must be an integer, we consider \( r = 49 \) and \( r = 50 \).

5. **Verifying the Maximum**:
We can verify that \( \binom{99}{49} = \binom{99}{50} \) because:
\[ \binom{n}{r} = \binom{n}{n-r} \]
Therefore:
\[ \binom{99}{49} = \binom{99}{50} \]

6. **Conclusion**:
The values of \( r \) for which \( P(X = r) \) is maximized are \( 49 \) and \( 50 \).

### Answer:
The correct option is:
**a. 49, 50**