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Question
Write the values of 'a' for which the following distribution of probabilities becomes a probability distribution:
| X= xi: | -2 | -1 | 0 | 1 |
| P(X= xi) : |
\[\frac{1 - a}{4}\]
|
\[\frac{1 + 2a}{4}\]
|
\[\frac{1 - 2a}{4}\]
|
\[\frac{1 + a}{4}\]
|
Solution
We know that the sum of probabilities in a probability distribution is always 1.
∴ P (X =-2) + P (X =-1) + P (X = 0) + P (X = 1) = 1
\[ \Rightarrow \frac{4}{4} = 1\]
\[ \Rightarrow 1 = 1\]
\[\text{ Now } , \]
\[0 \leq \frac{1 - a}{4} \leq 1\]
\[ \Rightarrow 0 \leq 1 - a \leq 4\]
\[ \Rightarrow - 1 \leq - a \leq 3\]
\[ \Rightarrow - 3 \leq a \leq 1 ....... . . \left( 1 \right)\]
\[0 \leq \frac{1 + a}{4} \leq 1\]
\[ \Rightarrow 0 \leq 1 + a \leq 4\]
\[ \Rightarrow - 1 \leq a \leq 3 ...... . . \left( 2 \right)\]
\[0 \leq \frac{1 - 2a}{4} \leq 1\]
\[ \Rightarrow 0 \leq 1 - 2a \leq 4\]
\[ \Rightarrow - 1 \leq - 2a \leq 3\]
\[ \Rightarrow - \frac{3}{2} \leq a \leq \frac{1}{2} ..... . . \left( 3 \right)\]
\[0 \leq \frac{1 + 2a}{4} \leq 1\]
\[ \Rightarrow 0 \leq 1 + 2a \leq 4\]
\[ \Rightarrow - 1 \leq 2a \leq 3\]
\[ \Rightarrow - \frac{1}{2} \leq a \leq \frac{3}{2} ...... . . \left( 4 \right)\]
\[\text{ From } \left( 1 \right), \left( 2 \right), \left( 3 \right) \text{ and } \left( 4 \right), \text{ we get} \]
\[ - \frac{1}{2} \leq a \leq \frac{1}{2}\]
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