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.
Answer & explanation
Correct answer: option 1
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 = $1-P(0)=1-q^5$
$=1-(\frac{4}{5})^5=1-\frac{1024}{3125}=\frac{2101}{3125}$