A fair coin is tossed 99 times. Let x be the number of times heads occurs. Then `P(X=r)` is maximum when r is .

Options

A. 49
B. 52
C. 51
D. none of these

Answer

To solve the problem, we need to determine the value 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.
- Let \( X \) be the number of times heads occurs.
- We need to find \( r \) such that \( P(X = r) \) is maximized.

2. **Identifying the Distribution**:
- Since the coin is fair, the probability of getting heads (success) is \( p = \frac{1}{2} \) and the probability of getting tails (failure) is \( q = \frac{1}{2} \).
- The number of trials \( n = 99 \).

3. **Using the Binomial Probability Formula**:
- The probability of getting exactly \( r \) heads in 99 tosses is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
- Substituting the values:
\[ P(X = r) = \binom{99}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{99 - r} = \binom{99}{r} \left(\frac{1}{2}\right)^{99} \]

4. **Maximizing the Probability**:
- To maximize \( P(X = r) \), we need to maximize \( \binom{99}{r} \) since \( \left(\frac{1}{2}\right)^{99} \) is a constant factor.
- The binomial coefficient \( \binom{n}{r} \) is maximized when \( r \) is close to \( \frac{n}{2} \).

5. **Calculating the Maximum Value of \( r \)**:
- For \( n = 99 \):
\[ r = \frac{99}{2} = 49.5 \]
- Since \( r \) must be an integer, we check \( r = 49 \) and \( r = 50 \).
- Both \( \binom{99}{49} \) and \( \binom{99}{50} \) will yield the maximum value.

6. **Conclusion**:
- Therefore, the value of \( r \) for which \( P(X = r) \) is maximum is either \( 49 \) or \( 50 \). However, since the question asks for a single value, we can state:
\[ r = 49 \]

### Final Answer:
The value of \( r \) for which \( P(X = r) \) is maximum is **49**.