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.
$i . e . \sum P (X) = 1$
$0 + k + 2 k + 2 k + 3 k + k 62 + 2 k 62 + 7 k^{2} + k = 1$
$10 k^{2} + 9 k - 1 = 0$
$10 k^{2} + 10 k - k - 1 = 0$
$(10 k - 1) (k + 1) = 0$
$k = \frac{1}{10} o r - 1$
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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