Probability Distribution of Random Variable X
X 1 2 3 2λ 3λ 4λ P(X) \(\frac{11}{30}\) \(\frac{1}{15}\) \(\frac{10}{30}\) \(\frac{3\lambda}{10}\) \(\frac{1}{15}\) \(\frac{1}{10}\)
(i) Calculate \( \lambda \), if \( E(X) = 3.2 \)
(ii) Find P(X > 1).
Probability Distribution of Random Variable X
| X | 1 | 2 | 3 | 2λ | 3λ | 4λ |
|---|---|---|---|---|---|---|
| P(X) | \(\frac{11}{30}\) | \(\frac{1}{15}\) | \(\frac{10}{30}\) | \(\frac{3\lambda}{10}\) | \(\frac{1}{15}\) | \(\frac{1}{10}\) |
(i) Calculate \( \lambda \), if \( E(X) = 3.2 \)
(ii) Find P(X > 1).
Solution and Explanation
(i) Find \( \lambda \) using total probability
We use the fact that the total sum of all probabilities must equal 1:
\[ \frac{11}{30} + \frac{1}{15} + \frac{10}{30} + \frac{3\lambda}{10} + \frac{1}{15} + \frac{1}{10} = 1 \]
Convert all terms to have a denominator of 30:
\[ \frac{11}{30} + \frac{2}{30} + \frac{10}{30} + \frac{9\lambda}{30} + \frac{2}{30} + \frac{3}{30} = 1 \]
Simplify the constants:
\[ \frac{28 + 9\lambda}{30} = 1 \]
Multiply both sides by 30:
\[ 28 + 9\lambda = 30 \Rightarrow 9\lambda = 2 \Rightarrow \lambda = \frac{2}{9} \]
Therefore, \( \lambda = \frac{2}{9} \)
(ii) Find \( P(X > 1) \)
This corresponds to \( X = 2, 3, 2\lambda, 3\lambda, 4\lambda \):
\[ P(X > 1) = P(X=2) + P(X=3) + P(X=2\lambda) + P(X=3\lambda) + P(X=4\lambda) \]
Substitute the values:
\[ P(X > 1) = \frac{1}{15} + \frac{10}{30} + \frac{3 \times \lambda}{10} + \frac{1}{15} + \frac{1}{10} \]
Now substitute \( \lambda = \frac{2}{9} \):
\[ P(X > 1) = \frac{1}{15} + \frac{1}{3} + \frac{6}{30} + \frac{1}{15} + \frac{1}{10} \]
Convert to common denominator (LCM = 30):
\[ P(X > 1) = \frac{2}{30} + \frac{10}{30} + \frac{6}{30} + \frac{2}{30} + \frac{3}{30} = \frac{23}{30} \]
Therefore, \( P(X > 1) = \frac{23}{30} \)
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