Given n = 12
P = 0.9
(i) P(X = 10) = 12C₁₀(0.9)¹⁰(1 - 0.9)²
= 12C₁₀(0.9)¹⁰ (0.1)²
(ii) P(X 2: 11) = R(X = 11) + P(X = 12)
= 12C₁₁(0.9)¹¹ (0.1) + 12C₁₂(0.9)¹²(0.1)⁶
= 12C (0.9)¹¹ (0.1) + (0.9)¹²
= 12(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)11 (1.2 + 0.9) = (0.9)11 (2.1)
(iii) P( at least 2 will not have a useful life of atleast 600 hours) = 1 - P( atleast 11 will have a useful life of atleast 600 hours)
= 1 - P(X ≥11)
= 1-2.1 (0.9)¹¹

\( \mathrm{F}(\mathrm{x}) =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right) \)
\(\mathrm{P}(0 \leq \mathrm{x} \leq 1) =\mathrm{F}(1)-\mathrm{F}(0) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+0\right) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}-\frac{\pi}{2}\right)=\frac{1}{4}\)

The probability distribution

\( \mathrm{E}(\mathrm{X}) =\sum p \mathrm{x} \)
\( =1 \cdot \mathrm{p}+2 \cdot \mathrm{qp}+3 \cdot \mathrm{q}^{2} \mathrm{p}+\ldots .+\mathrm{n} \mathrm{q}^{\mathrm{n}-1} \mathrm{p}+\ldots \)
\(=\mathrm{p}\left(1+2 \mathrm{q}+3 \mathrm{q}^{2}+\ldots . .+\mathrm{nq}^{\mathrm{n}-1}+\ldots \ldots\right) \)
\( =\mathrm{p}(1-\mathrm{q})^{-2} \)
\( =\mathrm{p}(\mathrm{p})^{-2}=\frac{p}{p^{2}}=\frac{1}{p} \) 

Given, np = 6
\( \sqrt{n p q} =3 \)
\(\mathrm{npq} =9 \)
\(\frac{n p q}{n p} =\frac{9}{6} \)
\(\mathrm{q}=\frac{3}{2}>1, \) (not possible)
Since p,q < 1 and p+q = 1
Therefore the given statement is wrong 

Let m + 1 and m be the mean and variance
Given, \((m+1)^{2}-m^{2}=11\)
\( m^{2}+2 m+1-m^{2} =11 \)
2 m = 10  
m = 5 
Mean = m + 1 = 6
np = 6, npq = 5, \(q=\frac{5}{6}, p=\frac{1}{6}\)
n = 36

Given, np + npq = 4.8
np(1 + q) = 4.8
5p(1 +(1-P)) = 4.8
\( 5 p^{2}-10 p+4.8 =0 \)
\(\left(p-\frac{6}{5}\right)\left(p-\frac{4}{5}\right) =0 \)
\(p =\frac{6}{5}, \frac{4}{5} \)
\(p =\frac{4}{5} \)
[\(\therefore\) p cannot greater than 1]
\(q=\frac{1}{5}\)
Binomial distribution is P(X = x)
\(={ }^{5} C_{x}\left(\frac{4}{5}\right)^{x}\left(\frac{1}{5}\right)^{5-x}, \mathrm{x}=0,1,2,3,4,5\)