Question
A random variable X has the following probability distribution:
Image
Determine:
(i) k
(ii) P(X < 3) (iii) P(X > 4)

Answer

(i) Since $P(x)$ is a probability distribution of $x$,
$ \sum_{x=0}^7 P(x)=1$
$\Rightarrow P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)=1$
$\Rightarrow 0+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+7 \mathrm{k}^2+\mathrm{k}=1$
$\Rightarrow 10 \mathrm{k}^2+9 \mathrm{k}-1=0$
$\Rightarrow 10 \mathrm{k}^2+10 \mathrm{k}-\mathrm{k}-1=0$
$\Rightarrow 10 \mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0$
$\Rightarrow(\mathrm{k}+1)(10 \mathrm{k}-1)=0$
$\Rightarrow 10 \mathrm{k}-1=0 \ldots \ldots[\because \mathrm{k} \neq-1]$
$\Rightarrow \mathrm{k}=\frac{1}{10} $
$ \text { (ii) } P(X<3)=P(0)+P(1)+P(2)$
$=0+k+2 k$
$=3 k$
$=3\left(\frac{1}{10}\right)$
$=\frac{3}{10} $
$ \text { (iii) } P(0=k+2 k$
$=3 k$
$=3\left(\frac{1}{10}\right)$
$=\frac{3}{10} $

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