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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.
Solution
Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)
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
\[ \Rightarrow 3P\left( X = - 1 \right) = k ................ \left [ \because P\left( X = - 1 \right) = P\left( X = - 2 \right) = P\left( X = - 3 \right) \right]\]
\[ \Rightarrow P\left( X = - 1 \right) = \frac{k}{3}\]
\[ \Rightarrow P\left( X = - 1 \right) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\]
\[ \therefore P\left( X = - 1 \right) = P\left( X = - 2 \right) = P\left( X = - 3 \right) = \frac{1}{9}\]
\[\text{ Similarly,} \]
\[P\left( X > 0 \right) = k\]
\[ \Rightarrow P\left( X = 1 \right) = P\left( X = 2 \right) = P\left( X = 3 \right) = \frac{1}{9}\]
| Xi | Pi |
| -3 |
\[\frac{1}{9}\]
|
| -2 |
\[\frac{1}{9}\]
|
| -1 |
\[\frac{1}{9}\]
|
| 1 |
\[\frac{1}{9}\]
|
| 2 |
\[\frac{1}{9}\]
|
| 3 |
\[\frac{1}{9}\]
|
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