A fair coin is tossed ` n` times and ` x` be the number of heads. If ` P(x=4), P(x=5), P(x=6)` are in A.P. then which of the following are correct.
Statement-I : ` n = 7` Statement-II : ` n = 9`
Statement-III : ` n = 2` Statement-IV : ` n = 14`
correct statements are

Options

A. only I
B. both I and III
C. both III and IV
D. both I and IV

Answer

To solve the problem, we need to analyze the probabilities of getting a specific number of heads when a fair coin is tossed `n` times. The probabilities of getting `x` heads in `n` tosses can be expressed using the binomial distribution formula:

\[ P(X = r) = \binom{n}{r} \left(\frac{1}{2}\right)^n \]

where \( \binom{n}{r} \) is the binomial coefficient, which counts the number of ways to choose `r` successes (heads) in `n` trials (tosses).

Given that \( P(X = 4), P(X = 5), P(X = 6) \) are in Arithmetic Progression (A.P.), we can set up the equation:

\[ 2P(X = 5) = P(X = 4) + P(X = 6) \]

Substituting the probabilities into this equation, we have:

\[ 2 \cdot \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{4} \left(\frac{1}{2}\right)^n + \binom{n}{6} \left(\frac{1}{2}\right)^n \]

We can simplify this by canceling \( \left(\frac{1}{2}\right)^n \) from both sides:

\[ 2 \cdot \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

Now, we can express the binomial coefficients in terms of factorials:

\[ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]

Substituting these into the equation gives:

\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]

Dividing through by \( n! \) (assuming \( n! \neq 0 \)):

\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

Next, we can multiply through by \( 5!(n-5)! \) to eliminate the denominators:

\[ 2 = \frac{5!}{4!} \cdot \frac{(n-5)!}{(n-4)!} + \frac{5!}{6!} \cdot \frac{(n-5)!}{(n-6)!} \]

This simplifies to:

\[ 2 = 5 \cdot \frac{1}{n-4} + \frac{1}{6} \cdot (n-5) \]

Now, multiplying through by \( 6(n-4) \):

\[ 12(n-4) = 30 + (n-5)(n-4) \]

Expanding both sides:

\[ 12n - 48 = 30 + n^2 - 9n + 20 \]

Rearranging gives:

\[ n^2 - 21n + 98 = 0 \]

Now we can factor this quadratic equation:

\[ (n - 7)(n - 14) = 0 \]

Thus, the solutions for \( n \) are:

\[ n = 7 \quad \text{or} \quad n = 14 \]

Now, we check the statements:

- Statement-I: \( n = 7 \) (True)
- Statement-II: \( n = 9 \) (False)
- Statement-III: \( n = 2 \) (False)
- Statement-IV: \( n = 14 \) (True)

Thus, the correct statements are Statement-I and Statement-IV.