If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

Concept:

Binomial distribution The binomial distribution formula is:P (X = r) = \({{\bf{n}}_{{{\bf{C}}_{\bf{r}}}}}{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}\)

P(X) = probability of a success on an individual trial

p = probability of “success”

r = number of success

q = probability of “failure” OR q = 1- p

n = number of trials

• \({{\rm{n}}_{{{\rm{C}}_{\rm{r}}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}\)

• Mean = np
• Variance = npq

Calculation:

Given: mean = 2, variance = 1

\(\frac{{{\rm{mean}}}}{{{\rm{varaince}}}} = \frac{{{\rm{np}}}}{{{\rm{npq}}}} = \frac{2}{1}\)

\(\Rightarrow {\rm{q}} = \frac{1}{2}\)

We know p + q = 1

So, p = 1 - ½ = ½

mean = 2

⇒ np = 2

⇒ n = 2/p

= 4

We know P(required) = 1 – P (not required)

P (X > 1) = 1 – P (X = 0) – P (X = 1)

\( \Rightarrow 1 - {4_{{{\rm{C}}_0}}}{\left( {\frac{1}{2}} \right)^0}{\left( {\frac{1}{2}} \right)^4} - {4_{{{\rm{C}}_1}}}{\left( {\frac{1}{2}} \right)^1}{\left( {\frac{1}{2}} \right)^3}\)    (P (X = r) = \({{\rm{n}}_{{{\rm{C}}_{\rm{r}}}}}{{\rm{p}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\))

\( \Rightarrow 1 - \left( 1 \right)\left( 1 \right){\left( {\frac{1}{2}} \right)^4} - \left( 4 \right)\left( {\frac{1}{2}} \right){\left( {\frac{1}{2}} \right)^3}\)         (∵ \({{\rm{n}}_{{{\rm{C}}_0}}} = 1{\rm{\;and\;}}{{\rm{n}}_{{{\rm{C}}_1}}} = {\rm{n}}\) )

\( \Rightarrow 1 - \left( {\frac{1}{{16}}} \right) - \frac{4}{{16}}\)

\(\Rightarrow 1 - \left( {\frac{5}{{16}}} \right)\)

\(\Rightarrow \left( {\frac{{11}}{{16}}} \right)\)

Hence, option (4) is correct