A random variable X has the following probability distribution XP(X=x)XP(X=x)0&#

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A random variable X has the following probability distribution

\begin{matrix} X & P (X = x) & X & P (X = x) 0 & λ & 5 & 11 λ 1 & 3 λ & 6 & 13 λ 2 & 5 λ & 7 & 15 λ 3 & 7 λ & 8 & 17 λ 4 & 9 λ \end{matrix}

then,

λ

is equal to

Options:

A . 181

B . 281

C . 581

D . 781

Answer: Option A
:
A

\sum_{x = 0}^{8} P (X) = 1 \Rightarrow P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) + P (7) + P (8) = 1 = λ + 3 λ + 5 λ + 7 λ + 9 λ + 11 λ + 13 λ + 15 λ + 17 λ = 1 \Rightarrow \frac{9}{2} (λ + 17 λ) = 1 \Rightarrow 81 λ = 1 ∴ λ = \frac{1}{81}

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