Question

A random variable X has following p.d.f. \( f(x) = kx(1 - x), 0 \le x \le 1 \) and \( P(x>a) = \frac{20}{27} \), then \( a = \)}

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{4} \)

Hint:

Always normalize the p.d.f. first by ensuring the total area under the curve is 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
First, find the normalization constant \( k \). Then use the given probability to set up an integral and solve for \( a \).
Step 2: Key Formula or Approach:
1. Total probability: \( \int_{0}^{1} f(x) dx = 1 \).
2. \( P(X>a) = \int_{a}^{1} f(x) dx = \frac{20}{27} \).
Step 3: Detailed Explanation:
Finding \( k \):
\[ \int_{0}^{1} k(x - x^2) dx = k \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = k \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{k}{6} = 1 \implies k = 6 \] Now use \( P(X>a) = \frac{20}{27} \):
\[ \int_{a}^{1} 6(x - x^2) dx = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_a^1 = 6 \left[ \left(\frac{1}{6}\right) - \left(\frac{a^2}{2} - \frac{a^3}{3}\right) \right] = 1 - 3a^2 + 2a^3 = \frac{20}{27} \] \[ 2a^3 - 3a^2 + 1 - \frac{20}{27} = 0 \implies 2a^3 - 3a^2 + \frac{7}{27} = 0 \] Testing \( a = \frac{1}{3} \):
\[ 2\left(\frac{1}{27}\right) - 3\left(\frac{1}{9}\right) + \frac{7}{27} = \frac{2}{27} - \frac{9}{27} + \frac{7}{27} = 0 \] So, \( a = \frac{1}{3} \).
Step 4: Final Answer:
The value of \( a \) is \( \frac{1}{3} \).