Question

If the function defined by $f(x) = K(x - x^2)$ if $0 < x < 1$ and $f(x) = 0$ otherwise, is the probability density function (p.d.f.) of a random variable X, then the value of $P\left(X < \frac{1}{2}\right)$ is

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

Hint:

Look for symmetry to save time! The quadratic polynomial $f(x) = K(x - x^2)$ represents a symmetric parabola centered exactly at its midpoint, $x = \frac{1}{2}$. Since a probability density function is completely symmetric around its mean line, the total area must split perfectly in half, meaning the probability of landing below $\frac{1}{2}$ is automatically $\frac{1}{2}$ without doing any integration at all!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a piecewise continuous function containing an unknown normalization constant $K$. Since this function represents a valid probability density function (p.d.f.), we first use the total probability condition to find $K$, and then integrate the function over the restricted interval to evaluate $P\left(X < \frac{1}{2}\right)$.

Step 2: Key Formula or Approach:
For any valid probability density function, the total area enclosed under the curve across the entire real domain must equal exactly 1:
$$\int_{-\infty}^{\infty} f(x) \, dx = 1 \implies \int_{0}^{1} K(x - x^2) \, dx = 1$$ Once $K$ is determined, the probability statement is found by calculating:
$$P\left(X < \frac{1}{2}\right) = \int_{0}^{1/2} f(x) \, dx$$

Step 3: Detailed Explanation:
1. Let's find the value of the constant $K$ using the normalization condition:
$$\int_{0}^{1} K(x - x^2) \, dx = 1 \implies K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1$$ Substitute the integration bounds:
$$K \left( \left(\frac{1}{2} - \frac{1}{3}\right) - 0 \right) = 1$$ $$K \left( \frac{3 - 2}{6} \right) = 1 \implies K\left(\frac{1}{6}\right) = 1 \implies K = 6$$ So our function is fully defined as $f(x) = 6(x - x^2)$ for $0 < x < 1$.
2. Now, compute the target probability accumulation from $x = 0$ to $x = \frac{1}{2}$:
$$P\left(X < \frac{1}{2}\right) = \int_{0}^{1/2} 6(x - x^2) \, dx = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1/2}$$ Distribute the constant multiplier 6 inside the brackets to simplify the fraction coefficients:
$$P\left(X < \frac{1}{2}\right) = \left[ 3x^2 - 2x^3 \right]_{0}^{1/2}$$ Substitute the upper bound value $x = \frac{1}{2}$:
$$P\left(X < \frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^3 = 3\left(\frac{1}{4}\right) - 2\left(\frac{1}{8}\right)$$ $$= \frac{3}{4} - \frac{2}{8} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$ This matches option (A).

Step 4: Final Answer:
The value of the probability $P\left(X < \frac{1}{2}\right)$ is $\frac{1}{2}$, which corresponds to option (A).