If the probability function of a random variable X is defined by … - Vidyadip

MCQ

If the probability function of a random variable $X$ is defined by $P(X=k)=a\left(\frac{k+1}{2^k}\right)$ for $k=0,1,2,3,4,5$, then the probability that X takes a prime value is

A
$\frac{13}{20}$
✓
$\frac{23}{60}$
C
$\frac{11}{20}$
D
$\frac{19}{60}$

✓

Answer

Correct option: B.

$\frac{23}{60}$

(B)

X = k	0	1	2	3	4	5
P(X = k)	a	a	$\frac{3a}{4}$	$\frac{4a}{8}$	$\frac{5a}{16}$	$\frac{6a}{32}$

$\begin{aligned}& \text { Since } \sum_{\mathrm{k}=0}^5 \mathrm{P}(\mathrm{X}=\mathrm{k})=1, \\& \mathrm{a}+\mathrm{a}+\frac{3 \mathrm{a}}{4}+\frac{4 \mathrm{a}}{8}+\frac{5 \mathrm{a}}{16}+\frac{6 \mathrm{a}}{32}=1 \\& \Rightarrow \frac{15}{4} \mathrm{a}=1 \Rightarrow \mathrm{a}=\frac{4}{15}\end{aligned}$
Now, $\mathrm{P}(\mathrm{X}=$ prime value $)$
$\begin{aligned}& =P(X=2)+P(X=3)+P(X=5) \\& =\frac{3 a}{4}+\frac{4 a}{8}+\frac{6 a}{32} \\& =\frac{23 a}{16} \\& =\frac{23}{16} \times \frac{4}{15}=\frac{23}{60}\end{aligned}$

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