Question

A random variable has the following probability distribution:

X = xi :	0	1	2	3	4	5	6	7
P (X = xi) :	0	2 p	2 p	3 p	p2	2 p2	7 p2	2 p

The value of p is

Options

•  1/10

• −1

• −1/10

• 1/5

   

Answer 1

 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

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