A random variable has the following probability distribution:

X = x_i : 0 1 2 3 4 5 6 7

P (X = x_i) : 0 2 p 2 p 3 p p² 2 p² 7 p² 2 p

The value of p is

(a) 1/10
(b) −1
(c) −1/10
(d) 1/5

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Solution

(a) 1/10

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

$\Rightarrow 0 + 2 p + 2 p + 3 p + p^{2} + 2 p^{2} + 7 p^{2} + 2 p = 1 \Rightarrow 10 p^{2} + 9 p - 1 = 0 \Rightarrow (10 p - 1) (p + 1) = 0 \Rightarrow p = \frac{1}{10} or - 1 (Neglecting - 1 as the value of the probabilitiy cannot be negative)$

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