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Definite Integral as Limit of Sum

The range of ...

Question

The range of a random variable $X$ is ${0, 1, 2}$ and $P (X = 0) = 3 K^{3}, P (X = 1) = 4 K - 10 K^{2}$
$P (x = 2) = 5 K - 1$ . Then we have

P (X = 0) < P (X = 2) < P (X = 1)

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P (X = 0) < P (X = 1) < P (X = 2)

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P (X = 1) + P (X = 0) = P (X = 2)

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P (X = 1) > P (X = 0) + P (X = 2)

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Solution

The correct option is B $P (X = 0) < P (X = 1) < P (X = 2)$
Using $\sum P (X) = 1$

$\Rightarrow 3 K^{3} - 10 K^{2} + 4 k + 5 K - 1 = 3 K^{3} - 10 K^{2} + 9 k - 1 = 1 \Rightarrow 3 K^{3} - 10 K^{2} + 9 K - 2 = 0$

Solving this we get $K = \frac{1}{3}, 1, 2$ but $0 < K < 1$ so

$K = \frac{1}{3}$

Thus $P (X = 0) = \frac{1}{9}, P (X = 1) = \frac{2}{9}, P (X = 2) = \frac{2}{3}$

Hence order is $P (X = 0) < P (X = 1) < P (X = 2)$

Suggest Corrections

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