If the sum and the product of the mean and variance of Binomial Distribution are 1.8 and 0.8 respectively, find the probability distribution and the probability of atleast one success.

$\frac{2101}{3125}$

The correct answer is Option (1) → $\frac{2101}{3125}$

According to given, we have

$np + npq = 1.8 ⇒ np(1 + q)=\frac{9}{5}$   ...(i)

and $np. npq=0.8⇒n^2p^2q= \frac{4}{5}$   ...(ii)

Dividing the square of (i) by (ii), we get

$\frac{n^2p^2(1 + q)^2}{n^2p^2q}=(\frac{9}{5})^2×\frac{5}{4}⇒\frac{(1 + q)^2}{q}=\frac{81}{20}$

$⇒20(1+2q + q^2) = 81q⇒20q^2 - 41q + 20 = 0$

$⇒(5q-4) (4q-5)= 0⇒q=\frac{4}{5},\frac{5}{4}$ but $0<q<1$

$∴p=1-q=1-\frac{4}{5}=\frac{1}{5}$

From (i), $n.\frac{1}{5}(1+\frac{4}{5})=\frac{9}{5}⇒n=5$

Hence, the binomial distribution is $(q+p)^n$ i.e. $\left(\frac{4}{5}+\frac{1}{5}\right)^5$

Probability of atleast one success = $1-P(0)=1-q^5$

$=1-(\frac{4}{5})^5=1-\frac{1024}{3125}=\frac{2101}{3125}$