An unbiased coin is tossed n times. Let X denote the number of times head occurs. If `P(X=4), P(X=5) and P(X=6)` are in A.P, then the value of n can be

Options

A. 7,14
B. 10,14
C. 12,7
D. none of these

Answer

To solve the problem, we need to find the value of \( n \) such that the probabilities \( P(X=4) \), \( P(X=5) \), and \( P(X=6) \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Binomial Distribution**:
Since we are tossing a coin \( n \) times, the number of heads \( X \) follows a binomial distribution with parameters \( n \) (number of trials) and \( p = \frac{1}{2} \) (probability of getting heads in a single toss). The probability mass function is given by:
\[ P(X = r) = \binom{n}{r} \left(\frac{1}{2}\right)^n \]
where \( \binom{n}{r} \) is the binomial coefficient.

2. **Setting Up the Probabilities**:
We need to express the probabilities for \( X = 4 \), \( X = 5 \), and \( X = 6 \):
\[ P(X=4) = \binom{n}{4} \left(\frac{1}{2}\right)^n \]
\[ P(X=5) = \binom{n}{5} \left(\frac{1}{2}\right)^n \]
\[ P(X=6) = \binom{n}{6} \left(\frac{1}{2}\right)^n \]

3. **Condition for Arithmetic Progression**:
The condition for \( P(X=4) \), \( P(X=5) \), and \( P(X=6) \) to be in AP is:
\[ 2P(X=5) = P(X=4) + P(X=6) \]

4. **Substituting the Probabilities**:
Substituting the expressions for the probabilities:
\[ 2 \cdot \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{4} \left(\frac{1}{2}\right)^n + \binom{n}{6} \left(\frac{1}{2}\right)^n \]
We can cancel \( \left(\frac{1}{2}\right)^n \) from both sides:
\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

5. **Using Binomial Coefficient Properties**:
We can express the binomial coefficients in terms of factorials:
\[ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]
Therefore, we can rewrite the equation:
\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]
Cancelling \( n! \) gives:
\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

6. **Finding a Common Denominator**:
Multiply through by \( 30(n-5)! \) (the least common multiple of the denominators):
\[ 12 = 6(n-4) + 5(n-6) \]
Expanding and simplifying:
\[ 12 = 6n - 24 + 5n - 30 \]
\[ 12 = 11n - 54 \]
\[ 11n = 66 \implies n = 6 \]

7. **Finding Possible Values of n**:
We need to check for other values of \( n \). The quadratic equation derived from the binomial coefficients gives:
\[ n^2 - 21n + 98 = 0 \]
Solving this using the quadratic formula:
\[ n = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 98}}{2 \cdot 1} = \frac{21 \pm \sqrt{441 - 392}}{2} = \frac{21 \pm 7}{2} \]
This yields:
\[ n = 14 \quad \text{or} \quad n = 7 \]

### Final Answer:
The possible values of \( n \) are \( 7 \) and \( 14 \).