# Let a random variable X have a binomial distribution with mean 8 and variance 4. If $P\left( {X \le 2} \right) = \frac{k}{{{2^{16}}}}$, then k is eq

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Let a random variable X have a binomial distribution with mean 8 and variance 4. If $P\left( {X \le 2} \right) = \frac{k}{{{2^{16}}}}$, then k is equal to:
1. 17
2. 121
3. 1
4. 137

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Correct Answer - Option 4 : 137

Given, the mean, μ = 8

and the variance is σ2 = 4

Let number of trials be n

Let probability of success be p

Let probability of failure be q

We know that, for Binomial distribution,

Mean μ = np,

⇒ np = 8

Variance, σ2 = npq

⇒ npq = 4

Now, the value of ‘q’ is:

⇒ (8)q = 4

$\Rightarrow q = \frac{4}{8}$

$\therefore q = \frac{1}{2}$

∵ [p + q = 1]

Now, the value of ‘p’ is:

$\Rightarrow {\rm{p}} + \frac{1}{2} = 1$

$\Rightarrow p = 1 - \frac{1}{2}$

$\therefore p = \frac{1}{2}$

Now, the value of ‘n’ is:

$\Rightarrow n\left( {\frac{1}{2}} \right) = 8$

∴ n = 16

Given, $P\left( {X \le 2} \right) = \frac{k}{{{2^{16}}}}$

$\Rightarrow P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right) = \frac{k}{{{2^{16}}}}$

$\Rightarrow P\left( {X \le 2} \right) = \frac{{{}_{\rm{\;}}^{16}{C_0} + {}_{\rm{\;}}^{16}{C_1} + {}_{\rm{\;}}^{16}{C_2}}}{{{2^{16}}}} = \frac{k}{{{2^{16}}}}$

${}_{\rm{\;}}^{16}{C_0} = \frac{{16!}}{{0!\left( {16 - 0} \right)!}} = \frac{{16!}}{{16!}}$

${}_{\rm{\;}}^{16}{C_1} = \frac{{16!}}{{1!\left( {16 - 1} \right)!}} = \frac{{16!}}{{15!}} = \frac{{16 \times 15!}}{{15!}} = 16$

${}_{\rm{}}^{16}{C_2} = \frac{{16!}}{{2!\left( {16 - 2} \right)!}} = \frac{{16!}}{{2! \times 14!}}$

$= \frac{{16 \times 15 \times 14!}}{{2 \times 14!}} = \frac{{16 \times 15}}{2} = 8 \times 15 = 120$

$\Rightarrow P\left( {X \le 2} \right) = \frac{{1 + 16 + 120}}{{{2^{16}}}} = \frac{k}{{{2^{16}}}}$

$\Rightarrow P\left( {X \le 2} \right) = \frac{{137}}{{{2^{16}}}} = \frac{k}{{{2^{16}}}}$

∴ k = 137