Question

If $f(x) = \begin{cases} 3(1 - 2x^2), & 0<x<1 \\ 0, & \text{otherwise} \end{cases}$ is a probability density function of $X$, then $P\left(\frac{1}{4}<x<\frac{1}{3}\right)$ is

• A. $\frac{75}{243}$
• B. $\frac{23}{96}$
• C. $\frac{179}{864}$
• D. $\frac{52}{243}$

Hint:

Probability Tip: When dealing with Probability Density Functions, always double-check that your requested interval lies entirely within the domain where the function rule is defined. If it crosses boundaries, you must split the integral into pieces.

Answer 1

The Correct Option is C

Concept: Probability - Continuous Random Variables and Probability Density Functions (PDF).

Step 1: Define the relationship between probability and PDF. For a continuous random variable $X$ with a probability density function $f(x)$, the probability that $X$ falls within a specific interval $[a, b]$ is given by the definite integral of $f(x)$ from $a$ to $b$: $P(a<X<b) = \int_{a}^{b} f(x) dx$.

Step 2: Set up the definite integral for the given interval. We need to calculate $P\left(\frac{1}{4}<X<\frac{1}{3}\right)$. Because the entire interval $\left(\frac{1}{4}, \frac{1}{3}\right)$ lies within the domain $(0, 1)$ where the function is defined as non-zero, we use $f(x) = 3(1 - 2x^2)$. The integral setup is: $P = \int_{1/4}^{1/3} 3(1 - 2x^2) dx$.

Step 3: Find the antiderivative of the function. First, expand the function inside the integral: $\int (3 - 6x^2) dx$. Apply the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$): $\int (3 - 6x^2) dx = 3x - \frac{6x^3}{3} = [3x - 2x^3]$.

Step 4: Evaluate the integral at the upper and lower limits. Evaluate at the upper limit ($x = \frac{1}{3}$): $3\left(\frac{1}{3}\right) - 2\left(\frac{1}{3}\right)^3 = 1 - 2\left(\frac{1}{27}\right) = 1 - \frac{2}{27} = \frac{25}{27}$. Evaluate at the lower limit ($x = \frac{1}{4}$): $3\left(\frac{1}{4}\right) - 2\left(\frac{1}{4}\right)^3 = \frac{3}{4} - 2\left(\frac{1}{64}\right) = \frac{3}{4} - \frac{1}{32}$. Make common denominators: $\frac{24}{32} - \frac{1}{32} = \frac{23}{32}$.

Step 5: Subtract to find the final probability. Apply the Fundamental Theorem of Calculus by subtracting the lower limit evaluation from the upper limit evaluation: $P = \frac{25}{27} - \frac{23}{32}$. Find a common denominator ($27 \times 32 = 864$): $P = \frac{25 \times 32}{864} - \frac{23 \times 27}{864} = \frac{800}{864} - \frac{621}{864} = \frac{179}{864}$. $$ \therefore \text{The required probability is } \frac{179}{864}. $$