Question

A random variable X takes the values 0, 1, 2 and 3 such that: 

P (X = 0) = P (X > 0) = P (X < 0); P (X = −3) = P (X = −2) = P (X = −1); P (X = 1) = P (X = 2) = P (X = 3) .  Obtain the probability distribution of X. 

Answer 1

Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)

\[\Rightarrow\] P (X > 0) = k
    P (X < 0) = k                         ∴ P (X = 0) + P (X > 0) + P (X < 0) = 1

\[\Rightarrow k + k + k = 1\]

\[ \Rightarrow k = \frac{1}{3}\]

Now,
P (X < 0) = k

\[\Rightarrow P\left( X = - 1 \right) + P\left( X = - 2 \right) + P\left( X = - 3 \right) = k\]
\[ \Rightarrow 3P\left( X = - 1 \right) = k   ................ \left [ \because P\left( X = - 1 \right) = P\left( X = - 2 \right) = P\left( X = - 3 \right) \right]\]
\[ \Rightarrow P\left( X = - 1 \right) = \frac{k}{3}\]
\[ \Rightarrow P\left( X = - 1 \right) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\]
\[ \therefore P\left( X = - 1 \right) = P\left( X = - 2 \right) = P\left( X = - 3 \right) = \frac{1}{9}\]
\[\text{ Similarly,}  \]
\[P\left( X > 0 \right) = k\]
\[ \Rightarrow P\left( X = 1 \right) = P\left( X = 2 \right) = P\left( X = 3 \right) = \frac{1}{9}\]

Thus, the probability distribution is given by

Xi	Pi
-3	\[\frac{1}{9}\]
-2	\[\frac{1}{9}\]
-1	\[\frac{1}{9}\]
1	\[\frac{1}{9}\]
2	\[\frac{1}{9}\]
3	\[\frac{1}{9}\]