Question

The diameter of an electric cable \(X\) is a continuous random variable with probability density function \[ f(x)=kx(1-x),\quad 0\leq x\leq 1. \] Find \[ P\left(X<\frac{1}{2}\ \middle|\ \frac{1}{3}<X<\frac{2}{3}\right) \]

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

Hint:

For conditional probability with continuous random variables, use: \[ P(A|B)=\frac{\int_{A\cap B} f(x)\,dx}{\int_B f(x)\,dx} \]

Answer 1

The Correct Option is B

Concept:
For conditional probability: \[ P(A|B)=\frac{P(A\cap B)}{P(B)} \] Here: \[ A:\ X<\frac{1}{2} \] and: \[ B:\ \frac{1}{3}<X<\frac{2}{3} \] So: \[ A\cap B:\ \frac{1}{3}<X<\frac{1}{2} \] Therefore: \[ P\left(X<\frac{1}{2}\ \middle|\ \frac{1}{3}<X<\frac{2}{3}\right) = \frac{P\left(\frac{1}{3}<X<\frac{1}{2}\right)} {P\left(\frac{1}{3}<X<\frac{2}{3}\right)} \]

Step 1: First find the value of \(k\).
Since \(f(x)\) is a probability density function: \[ \int_0^1 kx(1-x)\,dx=1 \] \[ k\int_0^1 (x-x^2)\,dx=1 \] \[ k\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=1 \] \[ k\left(\frac{1}{2}-\frac{1}{3}\right)=1 \] \[ k\left(\frac{1}{6}\right)=1 \] \[ k=6 \] So: \[ f(x)=6x(1-x) \]

Step 2: Find the numerator probability.
\[ P\left(\frac{1}{3}<X<\frac{1}{2}\right) = \int_{1/3}^{1/2}6x(1-x)\,dx \] \[ =\int_{1/3}^{1/2}6(x-x^2)\,dx \] \[ =\left[3x^2-2x^3\right]_{1/3}^{1/2} \] At \(x=\frac{1}{2}\): \[ 3\left(\frac{1}{2}\right)^2-2\left(\frac{1}{2}\right)^3 = \frac{3}{4}-\frac{1}{4} = \frac{1}{2} \] At \(x=\frac{1}{3}\): \[ 3\left(\frac{1}{3}\right)^2-2\left(\frac{1}{3}\right)^3 = \frac{1}{3}-\frac{2}{27} = \frac{9-2}{27} = \frac{7}{27} \] Therefore: \[ P\left(\frac{1}{3}<X<\frac{1}{2}\right) = \frac{1}{2}-\frac{7}{27} \] \[ =\frac{27-14}{54} \] \[ =\frac{13}{54} \]

Step 3: Find the denominator probability.
\[ P\left(\frac{1}{3}<X<\frac{2}{3}\right) = \int_{1/3}^{2/3}6x(1-x)\,dx \] \[ =\left[3x^2-2x^3\right]_{1/3}^{2/3} \] At \(x=\frac{2}{3}\): \[ 3\left(\frac{2}{3}\right)^2-2\left(\frac{2}{3}\right)^3 = 3\cdot\frac{4}{9}-2\cdot\frac{8}{27} \] \[ =\frac{4}{3}-\frac{16}{27} \] \[ =\frac{36-16}{27} \] \[ =\frac{20}{27} \] At \(x=\frac{1}{3}\): \[ 3\left(\frac{1}{3}\right)^2-2\left(\frac{1}{3}\right)^3 = \frac{7}{27} \] Therefore: \[ P\left(\frac{1}{3}<X<\frac{2}{3}\right) = \frac{20}{27}-\frac{7}{27} \] \[ =\frac{13}{27} \]

Step 4: Apply conditional probability formula.
\[ P\left(X<\frac{1}{2}\ \middle|\ \frac{1}{3}<X<\frac{2}{3}\right) = \frac{\frac{13}{54}}{\frac{13}{27}} \] \[ =\frac{13}{54}\times\frac{27}{13} \] \[ =\frac{27}{54} \] \[ =\frac{1}{2} \] Hence, the correct answer is: \[ \boxed{(B)\ \frac{1}{2}} \]