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Multiplication Rule

A random vari...

Question

$A random variable X has following probability distributions : The probability P (0 < X < 3) is$

X 0 1 2 3 4 5 6 7

P(X) 0 k 2k 2k 3k $k^{2}$ $2 k^{2}$ $7 k^{2} + k$

0.3

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Solution

The correct option is A 0.3
$We know that, the sum of a probability distribution of random variable is one. i . e ., \sum P (X) = 1 0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + 7 k^{2} + k = 1 10 k^{2} + 9 k - 1 = 0 10 k^{2} + 10 k - k - 1 = 0 (10 k - 1) (k + 1) = 0 k = \frac{1}{10} o r - 1 But k = -1 is rejected because probability cannot be negative ∴ k = \frac{1}{10} P (0 < X < 3) = P (X = 1) + P (X = 2) = k + 2 k = 3 k = \frac{3}{10} = 0.3$

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