A random variable ' X ' has the following probability distributio… - Vidyadip

MCQ

A random variable ' X ' has the following probability distribution

X	1	2	3	4	5	6	7
P(X)	k -1	3k	k	3k	$\mathrm{3k}^2$	$\mathrm{k}^2$	$\mathrm{k}^2+k$

Then the value of k is

A
-2
B
$\frac{1}{10}$
✓
$\frac{1}{5}$
D
$\frac{2}{7}$

✓

Answer

Correct option: C.

$\frac{1}{5}$

(C)
Since $\sum_{x=1}^7 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& \mathrm{k}-1+3 \mathrm{k}+\mathrm{k}+3 \mathrm{k}+3\mathrm{k}^2+\mathrm{k}^2+\mathrm{k}^2+\mathrm{k}=1 \\& \Rightarrow 5 \mathrm{k}^2+9 \mathrm{k}-2=0 \\& \Rightarrow 5 \mathrm{k}^2+10 \mathrm{k}-\mathrm{k}-2=0 \\& \Rightarrow 5 \mathrm{k}(\mathrm{k}+2)-1(\mathrm{k}+2)=0 \\& \Rightarrow(5 \mathrm{k}-1)(\mathrm{k}+2)=0 \\& \Rightarrow \mathrm{k}=\frac{1}{5} \quad \ldots[\because \mathrm{k}=-2 \text { is not possible }]\end{aligned}$

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