Question

In a binomial distribution, mean is 3 and varience is 3/2. Find the probability of at least 5 success ?

Answer 1

Since we have the mean$np=3$

Its variance$=npq=\frac{3}{2}$

$\Rightarrow 3\times q=\frac{3}{2}$

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

We know that $p+q=1$

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

Since we have $np=3,p=\frac{1}{2}$ we get $n\times \frac{1}{2}=3$ or $n=6$

Let us find the probability of $5$ successes by using Binomial Mass function

$P(X=x)={6_C}_5{\left(\frac{1}{2}\right)}^5\times \frac{1}{2}$

$=\frac{6!}{5!}\frac{1}{2^6}=\frac{6}{2^6}=\frac{3}{32}=0.0937$

$\therefore$ Probability of $5$ successes is $0.0937$