Which of the following statement is/are correct?
a) If a binomial distribution has mean 2.4 and variance is 1.44 then n=6
b) The probability that a student is not a swimmer is $\frac{1}{5}$ the probability that out of 5 students exactly 4 of are swimmers is ${\left(\frac{4}{5}\right)}^4$
c) In an  average rain falls on 12 days in every 30 days the probability that rain will fall on just 3 days a given week is $35{\left(\frac{2}{5}\right)}^3{\left(\frac{3}{5}\right)}^4$
d) One in 9 ships is likely  to be wrecked. When they are set on sail. When 6 ships set on sail, the probability for exactly 3 will not arrive safely is ${ }^6{C}_3\left(\frac{8^3}{9^6}\right)$

a) Given that np = 2.4 , npq = 1.44
$\begin{array}{l}q=\frac{1.44}{2.4}=315,p=2/5\\ \Rightarrow n\left(\frac{2}{5}\right)=2.4\\ \Rightarrow n=6\end{array}$ 
b) Now, the probability of students who are not swimmers 
$\begin{array}{l}q=1/5,p=1-q=\frac{4}{5}\\ andn=5\\ clearly,xhasabinomialdistributionwithn=5\\ P\left(x=r\right){=}^n{c}_r{p}^r{q}^{n-r}\\ P\left(x=4\right){=}^5{c}_4{\left(\frac{4}{5}\right)}^4{\left(\frac{1}{5}\right)}^{5-4}\\ ={\left(\frac{4}{5}\right)}^4\end{array}$ 
c) In an average rain falls on 12 days in energy 30 days
$\begin{array}{l}P=\frac{12}{30}=2/5,Q=1-\frac{2}{3}=\frac{3}{5}\\ Theprobabilitythatrainwillfallonjust3daysofagivenweek\\ n=7\left(week\right),r=3\\ P\left(X=r\right){=}^n{C}_r{p}^r{q}^{n-r}\\ P\left(X=3\right){=}^7{C}_3{\left(\frac{2}{3}\right)}^3{\left(\frac{3}{5}\right)}^{7-3}\\ =35{\left(\frac{2}{3}\right)}^3{\left(\frac{3}{5}\right)}^4\end{array}$ 
D) One in 9 ships is likely to be wrecked
$\begin{array}{l}P=\frac{1}{9},q=1-\frac{1}{9}=\frac{8}{9}\\ When6shipssetonsail,theprobabilityfor\end{array}$ 
$\begin{array}{l}exactly,3willnotarrivesafelyin\\ n=6,r=3\\ P\left(x=r\right){=}^n{C}_r{p}^r{q}^{n-r}\\ P\left(x=3\right){=}^6{C}_3{\left(\frac{1}{9}\right)}^3{\left(\frac{8}{9}\right)}^{6-3}\\ {=}^6{C}_3\left(\frac{8^3}{9^6}\right)\end{array}$