Question

Let X be a random variable with the probability density function defined by
\(f(x)=\frac{1}{x^{2}},\ 1\lt x\lt \infty\)
\(=0\) otherwise.
If \(C_{1}=\{x:1\lt x\lt 2\}\) and \(C_{2}=\{x:4\lt x\lt 5\}\) then \(P(C_{1}\cup C_{2})=\)

• A. \(0.36\)
• B. \(0.44\)
• C. \(0.63\)
• D. \(0.55\)

Hint:

The two intervals are disjoint, so add the integrals of the density over each.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For disjoint intervals, \(P(C_{1}\cup C_{2})=P(C_{1})+P(C_{2})\), with each probability being the integral of the density.

Step 2: Find each probability
\(P(C_{1})=\int_{1}^{2}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{1}^{2}=1-\frac{1}{2}=\frac{1}{2}\).
\(P(C_{2})=\int_{4}^{5}\frac{dx}{x^{2}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\).

Step 3: Add
\(\frac{1}{2}+\frac{1}{20}=\frac{11}{20}=0.55\).

Final Answer:
The probability is \(0.55\). \[ \boxed{\text{(D) }0.55} \]