Question

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).

Answer 1

(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, \( \lambda = \frac{2}{9} \)

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(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, \( P(X > 1) = \frac{23}{30} \)