Question

Determine the probability of $3$ successes in a binomial distribution whose mean and variance are respectively $2$ and $1.5$.

Answer 1

Let the parameters of the binomial distribution be $n,P$.

Give the mean is $2\Rightarrow np=2$

Variance is $1.5\Rightarrow npq=1.5$ where $q=1-p$

$q=\frac{1.5}{2}=\frac{3}{4}$

$P=1-q=1-\frac{3}{4}=\frac{1}{4}$

$np=2\Rightarrow n\times \frac{1}{4}=2\Rightarrow n=8$

The binomial distribution is ,

$p(x=r)={}^8{C}_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{8-r}$ for $r=0,1,......8.$

The probability of $3$ successes in the binomial distribution $=P(x=3)$

$={}^8{C}_3.{\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^5$

$=56.\frac{3^5}{4^8}$