Question

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

Answer 1

According to Question, we have
$np+npq=1.8$
$\Rightarrow np(1+q)=1.8$ ........ (i)
and $np.npq=0.8$
$\Rightarrow n^2{p}^{2}q=0.8$ ...... (ii)

Dividing the square of (i) by (ii) we get
$\frac{n^2{p}^2(1+q{)}^2}{n^2{p}^{2}q}=\frac{1.8\times 1.8}{0.8}$
$\Rightarrow \frac{(1+q{)}^2}{q}=\frac{3.24}{0.8}$
$\Rightarrow \frac{(1+q{)}^2}{q}=\frac{324}{80}$
$\Rightarrow \frac{(1+q{)}^2}{q}=\frac{81}{20}$
$\Rightarrow 20q^2-41q+20=0$
$\Rightarrow 20q^2-25q-16q+20=0$
$\Rightarrow 5q(4q-5)-4(4q-5)=0$
$(4q-5)(5q-4)=0$
$q=\frac{5}{4}$ or $q=\frac{4}{5}$ but $q\ne \frac{5}{4}$
$\therefore q=\frac{4}{5}$
$\Rightarrow p=\frac{1}{5}$

From (i), $n\times \frac{1}{5}(1+\frac{4}{5})=1.8$
$n\times \frac{1}{5}\times \frac{9}{5}=1.8$
$\Rightarrow 9n=25\times 1.8$
$\Rightarrow n=\frac{25\times 1.8}{9}$
$\Rightarrow n=\frac{45}{9}$
$\therefore n=5$

Hence, the binomial distribution is $(p+q{)}^n={(\frac{1}{5}+\frac{4}{5})}^5$
Probability of getting atleast 1 success $=P(r\ge 1)$
$=P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)$

${=}^5{C}_1.\frac{1}{5}{\left(\frac{4}{5}\right)}^4{+}^5{C}_2{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^3{+}^5{C}_3{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^2{+}^5{C}_4{\left(\frac{1}{5}\right)}^4\left(\frac{4}{5}\right){+}^5{C}_5{\left(\frac{1}{5}\right)}^5$

$=5.\frac{1}{5}\times \frac{256}{625}+10.\frac{1}{25}.\frac{64}{125}+10.\frac{1}{125}.\frac{16}{25}+5.\frac{1}{625}.\frac{4}{5}+1.\frac{1}{3125}$

$=\frac{256}{625}+\frac{128}{625}+\frac{32}{625}+\frac{4}{625}+\frac{1}{3125}$

$=\frac{1280+640+160+20+1}{3125}=\frac{2101}{3125}=0.67$