A random variable X has the following probability distribution- - X 1 2 3 4 5- PX K2 2K K 2K 5K2- -Then PX - 2 is equal to-

We know that ∑^5_x=1 P(X) = 1 ⇒ K^2 + 2K + K + 2K + 5K^2 = 1 ⇒ K = 1, 16⇒ K = 16 P(X gt; 2) = P(X = 3) + P(X = 4) + P(X = 5) = K + 2K + 5K^2 = 2336 ...

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Standard XII

Mathematics

Probability Distribution

A random vari...

Question

A random variable $X$ has the following probability distribution:
$\begin{matrix} X & 1 & 2 & 3 & 4 & 5 P (X) & K^{2} & 2 K & K & 2 K & 5 K^{2} \end{matrix}$
Then $P (X > 2)$ is equal to:

\frac{7}{12}

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\frac{23}{36}

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\frac{1}{36}

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\frac{1}{6}

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Solution

The correct option is B $\frac{23}{36}$
We know that $5 \sum x = 1 P (X) = 1$
$\Rightarrow K^{2} + 2 K + K + 2 K + 5 K^{2} = 1$
$\Rightarrow K = - 1, \frac{1}{6} \Rightarrow K = \frac{1}{6}$

$P (X > 2) = P (X = 3) + P (X = 4) + P (X = 5)$
$= K + 2 K + 5 K^{2} = \frac{23}{36}$

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A random variable X has the following probability distribution: X12345P(X)K22KK2K5K2 Then P(X>2) is equal to:

The correct option is B 2336We know that 5∑x=1P(X)=1 ⇒ K2+2K+K+2K+5K2=1 ⇒ K=−1,16⇒ K=16 P(X>2)=P(X=3)+P(X=4)+P(X=5) =K+2K+5K2=2336

A random variable $X$ has the following probability distribution:
$\begin{matrix} X & 1 & 2 & 3 & 4 & 5 P (X) & K^{2} & 2 K & K & 2 K & 5 K^{2} \end{matrix}$
Then $P (X > 2)$ is equal to:

The correct option is B $\frac{23}{36}$
We know that $5 \sum x = 1 P (X) = 1$
$\Rightarrow K^{2} + 2 K + K + 2 K + 5 K^{2} = 1$
$\Rightarrow K = - 1, \frac{1}{6} \Rightarrow K = \frac{1}{6}$

$P (X > 2) = P (X = 3) + P (X = 4) + P (X = 5)$
$= K + 2 K + 5 K^{2} = \frac{23}{36}$