A random variable $X$ has its range $X = 0, 1, 2$ with respective probabilities $P (X = 0) = 3 K^{3}, P (X = 1) = 4 K - 10 K^{2}, P (X = 2) = 5 K - 1$ , then the value of $K$ is

2

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1

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\frac{1}{3}

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2, 1, \frac{1}{3}

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Solution

The correct option is C $\frac{1}{3}$
Given, $P (X = 0) = 3 K^{3}, P (X = 1) = 4 K - 10 K^{2}, P (X = 2) = 5 K - 1$

Now, $P (X = 0) + P (X = 1) + P (X = 2) = 1$

$\Rightarrow 3 K^{3} - 10 K^{2} + 9 K - 2 = 0$

$\Rightarrow (K - 1) (3 K^{2} - 7 K + 2) = 0$

$\Rightarrow (K - 1) (K - 2) (3 K - 1) = 0$

$\Rightarrow K = 1, 2, \frac{1}{3}$

But $K = 1, 2$ not possible as at these values of $K$ , the probabilities $P (X = 0)$ , $P (X = 1)$ , $P (X = 2)$ does not lie between $0$ and $1$ .

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