A random variable x has the following probability distribution
The value of P is
x 0 1 2 3 4 5 6 7
p(x) 0 P 2P 2P 3P $p^{2}$ $2 P^{2}$ $7 P^{2} + P$

-1

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\frac{1}{10}

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- \frac{1}{10}

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None of these

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Solution

The correct option is C $\frac{1}{10}$
For a probability distribution,

$\sum_{i = 1}^{n} P_{i} = 1$

Therefore,
$0 + P + 2 P + 2 P + 3 P + P^{2} + 2 P^{2} + 7 P^{2} + P = 1$

$10 P^{2} + 9 P = 1$

$10 P^{2} + 9 P - 1 = 0$

$10 P^{2} + 10 P - P - 1 = 0$

$10 P (P + 1) - 1 (P + 1) = 0$

$(10 P - 1) (P + 1) = 0$

$P = \frac{1}{10}$ or $P = - 1$

Probability cannot be negative, so value of $P = \frac{1}{10}$

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