Question

In a binomial distribution, mean is $3$ and variance is $3/2$. Find the probability of at least $4$ success.

Answer 1

A] E(x)=np=3

v(x)=npq=$\frac{3}{2}$

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

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

np = 3

$\therefore np=n\times \frac{1}{2}=3\Rightarrow n=6$

$p(x=x){=}^n{c}_x{p}^x{q}^{n-x}$ $(xϵ64,6)$

P(X=4)+P(X=5)+P(X=6)

${}^6{C}_4\left(\frac{1}{2}{)}^4\right(\frac{1}{2}{)}^{6-4}+6\left(\frac{1}{2}{)}^5\right(\frac{1}{2}{)}^{6-5}{+}^6{C}_6(\frac{1}{2}{)}^6\frac{1}{2}$

$=\frac{6\times 5}{2}\times \frac{1}{2^6}+6\frac{1}{2^6}+1\times \frac{1}{2^6}$

$=\frac{1}{2^6}[15+6+1]=\frac{1}{2^6}\times 22$

$=\frac{11}{25}=\frac{11}{32}$