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# Let X Denote the Number of Times Heads Occur in N Tosses of a Fair Coin. If P (X = 4), P (X = 5) and P (X = 6) Are in Ap, the Value of N is (A) 7, 14 (B) 10, 14 (C) 12, 7 (D) 14, 12 - Mathematics

#### Question

Let X denote the number of times heads occur in n tosses of a fair coin. If P (X = 4), P (X= 5) and P (X = 6) are in AP, the value of n is

• 7, 14

• 10, 14

• 12, 7

• 14, 12

#### Solution

7, 14

$\text{ Here } , p = \frac{1}{2}\text{ and } q = \frac{1}{2}$
$\text{ Binomial distribution is given by }$
$P(X = r) =^{n}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{n - r}$

P (X = 4), (= 5), P(= 6) are in A.P.

$\therefore ^{n}{}{C}_4 +^{n}{}{C}_6 = 2 ^{n}{}{C}_5$

$\Rightarrow \frac{n(n - 1)(n - 2)(n - 3)}{2\left( 4! \right)} + \frac{n(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)}{2\left( 6! \right)} = \frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{5!}$

$\text{ By simplifying, we get}$

$\frac{1}{2} + \frac{(n - 4)(n - 5)}{2(30)} = \frac{n - 4}{5}$

$\text{ Taking LCM as 60, we get}$

$30 + n^2 - 9n + 20 = 12n - 48$

$\Rightarrow n^2 - 21n + 98 = 0$

$\Rightarrow (n - 7)(n - 14) = 0$

$\Rightarrow n = 7, 14$

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