Question

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

Answer 1

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