A random variable X has the following probability distributionWhelend array then- - is equal toA- 1-81B- 2-81C- 5-81D- 7-81

The correct option is A 1/81∑_x=0^8P(X)=1 ⇒ P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1 =λ+3 λ+5λ +7λ +9λ +11λ +13λ+15 λ+17 λ=1 ⇒9/2(λ+17λ)=1 ⇒ 81 λ =1 ∴λ= ...

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Standard XII

Mathematics

Probability Distribution

A random vari...

Question

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

\frac{1}{81}

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\frac{2}{81}

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\frac{5}{81}

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\frac{7}{81}

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Solution

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

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A random variable X has the following probability distribution XP(X=x)XP(X=x)0λ511λ13λ613λ25λ715λ37λ817λ49λ then, λ is equal to

The correct option is A 181∑8x=0P(X)=1⇒P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1=λ+3λ+5λ+7λ+9λ+11λ+13λ+15λ+17λ=1⇒92(λ+17λ)=1⇒81λ=1∴λ=181

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