Question
A computer installation has 10 terminals. Independently, the probability that anyone terminal will require attention during a week is 0.1. Find the probabilities that (i) 0 (ii) 1 (iii) 2 (iv) 3 or more, terminals will require attention during the next week.

Answer

Let $\mathrm{X}=$ number of terminals which required attention during a week.
$\mathrm{p}=$ probability that any terminal will require attention during a week
$\therefore p=0.1$ and $q=1-p=1-0.1=0.9$
Given: $\mathrm{n}=10$
$\therefore X \sim \mathrm{B}(10,0.1)$
The p.m.f. of $X$ is given by
$\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $\mathrm{p}(\mathrm{x})={ }^{10} C_x(0.1)^x(0.9)^{10-x}, \mathrm{x}=0,1,2, \ldots, 10$
(i) $\mathrm{P}$ (no terminal will require attention $)=\mathrm{P}(\mathrm{X}=0)$
$ =p(0)={ }^{10} \mathrm{C}_0(0.1)^0(0.9)^{10-0}$
$=1 \times 1 \times(0.9)^{10}=(0.9)^{10} $
Hence, the probability that no terminal requires attention $=(0.9)^{10}$
(ii) $\mathrm{P}(1$ terminal will require attention $)$
$ P(X=1)=p(1)={ }^{10} C_1(0.1)^1(0.9)^{10-1}$
$=10(0.1)(0.9)^9$
$=(1.0)(0.9)^9$
$ =(0.9)^9 $
Hence, the probability that 1 terminal requires attention $=(0.9)^9$.
(iii) $\mathrm{P}(2$ terminals will require attention)
$ P(X=2)=p(2)={ }^{10} \mathrm{C}_2(0.1)^2(0.9)^{10-2}$
$=\frac{10 \times 9}{1 \times 2}(0.1)^2(0.9)^8$
$=45(0.01)(0.9)^8$
$=(0.45) \times(0.9)^8 $
Hence, the probability that 2 terminals require attention $=(0.45)(0.9)^8$.
(iv) $\mathrm{P}$ (3 or more terminals will require attention)
$ =P(X \geqslant 3)$
$=1-P(x<3)$
$=1-[P(X=0)+P(X=1)+P(X=2)]$
$=1-[p(0)+p(1)+p(2)]$
$=1-\left[(0.9)^{10}+(0.9)^9+(0.45)(0.9)^8\right]$
$=1-\left[(0.9)^2+(0.9)^1+0.45\right](0.9)^8$
$=1-[0.81+0.9+0.45](0.9)^8$
$=1-(2.16) \times(0.9)^8 $
Hence, the probability that 3 or more terminals require attention $=1-(2.16) \times(0.9)^8$.

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