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Question
A random variable X has the following probability distribution:
| Values of X : | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P (X) : | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
Determine:
(i) The value of a
(ii) P (X < 3), P (X ≥ 3), P (0 < X < 5).
Solution
(i) Since the sum of probabilities in a probability distribution is always 1.
∴ P (X = 0) + 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
\[\Rightarrow a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1\]
\[ \Rightarrow 81a = 1\]
\[ \Rightarrow a = \frac{1}{81}\]
(ii) P (X < 3)
\[= P\left( X = 0 \right) + P\left( X = 1 \right) + P\left( X = 2 \right)\]
\[ = \frac{1}{81} + \frac{3}{81} + \frac{5}{81}\]
\[ = \frac{9}{81}\]
\[ = \frac{1}{9}\]
P (X ≥ 3)
\[P\left( X = 3 \right) + P\left( X = 4 \right) + P\left( X = 5 \right) + P\left( X = 6 \right) + P\left( X = 7 \right) + P\left( X = 8 \right)\]
\[ = \frac{7}{81} + \frac{9}{81} + \frac{11}{81} + \frac{13}{81} + \frac{15}{81} + \frac{17}{81}\]
\[ = \frac{72}{81}\]
\[ = \frac{8}{9}\]
P (0 < X < 5)
\[P\left( X = 1 \right) + P\left( X = 2 \right) + P\left( X = 3 \right) + P\left( X = 4 \right)\]
\[ = \frac{3}{81} + \frac{5}{81} + \frac{7}{81} + \frac{9}{81}\]
\[ = \frac{24}{81}\]
\[ = \frac{8}{27}\]
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