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

$$\begin{gathered} \Rightarrow k+k+k=1 \\ \Rightarrow k=\frac{1}{3} \end{gathered}$$

Now,
P (X < 0) = k

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

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	$\frac{1}{9}$