The range of a random variable X is {1,2,3…...} and P(X=x)=c^xx!, for x=1,2,3,…… Then the value of C is

Correct answer: (C) ln(2) (where ln denotes the natural log ). Given: PX=x=c^xx! ∵ ∑PX=1 ⇒ c+c^22!+c^33!+.....=1 ⇒ 1+c+c^22!+c^33!+...... =2 ⇒ e^c=2 ⇒ clog_ee=log_e2 ⇒c=ln2

Q79 / 160 · AP EAMCET 2021 (20 Aug Shift 2) · Mathematics

The range of a random variable $X$ is ${1, 2, 3 \dots . . .}$ and $P (X = x) = \frac{c^{x}}{x!}$ , for $x = 1, 2, 3, \dots \dots$ Then the value of $C$ is

A
$0$
B
$1$
Correct answer
C
$\ln (2) ($ where $\ln$ denotes the natural $\log$ )
D
$\ln (3) ($ where $\ln$ denotes the natural $\log$ )

Solution

Given: $P (X = x) = \frac{c^{x}}{x!}$
$∵ \sum_{}^{} P (X) = 1$
$\Rightarrow c + \frac{c^{2}}{2!} + \frac{c^{3}}{3!} + . . . . . = 1$
$\Rightarrow 1 + c + \frac{c^{2}}{2!} + \frac{c^{3}}{3!} + . . . . . . = 2$
$\Rightarrow e^{c} = 2$
$\Rightarrow c \log_{e} e = \log_{e} 2$
$\Rightarrow c = \ln 2$

Hence, the correct answer is option: 'C'.

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The range of a random variable X is {1,2,3…...} and P(X=x)=cxx!, for x=1,2,3,…… Then the value of C is

Solution

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The range of a random variable $X$ is ${1, 2, 3 \dots . . .}$ and $P (X = x) = \frac{c^{x}}{x!}$ , for $x = 1, 2, 3, \dots \dots$ Then the value of $C$ is