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: Find k
$\int_0^1 k(x - x^2) dx = 1 \implies k [x^2/2 - x^3/3]_0^1 = 1 \implies k/6 = 1 \implies k = 6$.
Step 2: Set up Probability Integral
$P(x>a) = \int_a^1 6(x - x^2) dx = \frac{20}{27}$.
$6 [x^2/2 - x^3/3]_a^1 = \frac{20}{27} \implies 6 [ (1/6) - (a^2/2 - a^3/3) ] = \frac{20}{27}$.
Step 3: Solve for a
$1 - 3a^2 + 2a^3 = \frac{20}{27} \implies 2a^3 - 3a^2 + \frac{7}{27} = 0$.
Testing $a = 1/3$: $2(1/27) - 3(1/9) + 7/27 = 2/27 - 9/27 + 7/27 = 0$.
Step 4: Conclusion
Hence, $a = 1/3$.
Final Answer:(A)