Maths Solve the Following Question.(2 Marks)

For X~ B(n, p), E(X) = np and V(X) = npq
Given that E(X) = 6 and V(X) = 4.2
$\begin{aligned} & \therefore \frac{V(X)}{E(X)}=\frac{4.2}{6} \\ & \therefore \frac{n p q}{n p}=0.7 \\ & \therefore q=0.7 \\ & \therefore p=1-q \\ & =1-0.7\end{aligned}$
= 0.3
∴ E(X) = np
= 6
⇒ n × 0.3 = 6
$\begin{aligned} & \Rightarrow n =\frac{6}{0.3} \\ & =20\end{aligned}$

Let r = no of bombs hit the target
p=0.8
q=0.2            (1-p=q)
n=10            r=4
$p(r=4)={ }^n C_r p^r q^{n-r} \quad r =0,1,2 \ldots \ldots \ldots, n$
$\begin{aligned} & ={ }^{10} C_4(0.8)^4(0.2)^6 \\ & ={ }^{10} C_4\left(\frac{8}{10}\right)^4\left(\frac{2}{10}\right)^6 \\ & =\frac{10 !}{4 ! 6 !} \times(2)^{18}\left(\frac{1}{10}\right)^{10} \\ & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2} \times(2)^{18} \times\left(\frac{1}{10}\right)^{10}\end{aligned}$
$\begin{aligned} & =210 \times(2)^{18} \times\left(\frac{1}{10}\right)^{10} \\ & =\frac{262144 \times 210}{(10)^{10}}=\frac{55050240}{(10)^{10}}\end{aligned}$
$\begin{aligned} & =\text { Anti }[\log 210+18 \log 2-10] \\ & =\text { Anti }[2.3222+18 \log (0.3010)-10] \\ & =\text { Anti }(3.7402) \\ & =0.0055\end{aligned}$

$\begin{aligned} & X \sim B(n=10, p) \\ & \therefore E(X)=n p \\ & 8=10 p \\ & p=0.8=\frac{4}{5}\end{aligned}$

Given, X ~ B (n, p) and E(X) = 6 and Var (X) = 4.2
$\begin{aligned} & \text { Now, } \frac{E(X)}{u(X)}=\frac{n p}{n p q} \\ & \Rightarrow \frac{6}{4.2}=\frac{1}{q} \\ & \Rightarrow q =\frac{4.2}{6}=0.7\end{aligned}$
Since,  p + q = 1
⇒ p + 0.7 = 1
⇒ p = 0.3
Now E(X) = np
⇒ 6 = n × 0.3
$\Rightarrow n =\frac{6}{0.3}=\frac{60}{3}=20$