Question:
Let X be a continuous random variable with the probability density function(p.d.f.) given by
\(f(x) = \left\{ \begin{array}{cc}kx, & 0\leq x < 1 \\ k, & 1\leq x < 2 \\ -kx+3k, & 2\leq x < 3 \\ 0, & \text{otherwise}\end{array} \right.\)
\(P(2 < X\leq 3) = \cdots\)
Let X be a continuous random variable with the probability density function(p.d.f.) given by
\(f(x) = \left\{ \begin{array}{cc}kx, & 0\leq x < 1 \\ k, & 1\leq x < 2 \\ -kx+3k, & 2\leq x < 3 \\ 0, & \text{otherwise}\end{array} \right.\)
\(P(2 < X\leq 3) = \cdots\)
\(f(x) = \left\{ \begin{array}{cc}kx, & 0\leq x < 1 \\ k, & 1\leq x < 2 \\ -kx+3k, & 2\leq x < 3 \\ 0, & \text{otherwise}\end{array} \right.\)
\(P(2 < X\leq 3) = \cdots\)
Show Hint
Total probability is 1, which fixes k; then integrate the last piece.
Updated On: Oct 1, 2026
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For a probability density function, the total area under the graph is 1. This fixes the constant \(k\).
Step 2: Find k:
\[ \int_0^1 kx\,dx + \int_1^2 k\,dx + \int_2^3 (-kx + 3k)\,dx = 1 \]
First piece: \(\frac k2\). Second piece: \(k\). Third piece: \(k\left[-\frac{x^2}{2} + 3x\right]_2^3 = k\left[\left(-\frac92 + 9\right) - \left(-2 + 6\right)\right] = k\left(\frac92 - 4\right) = \frac k2\).
Total: \(\frac k2 + k + \frac k2 = 2k = 1\), so \(k = \frac12\).
Step 3: Required probability:
\[ P(2 < X \le 3) = \int_2^3(-kx + 3k)\,dx = \frac k2 = \frac14 \]
Step 4: Why the other options are wrong.
\(\frac12\) is the value of \(k\), not the probability. \(\frac13\) and \(\frac15\) do not follow from the areas \(\frac14, \frac12, \frac14\).
Final Answer:
The probability is \(\frac14\), option (C). \[ \boxed{\frac{1}{4}} \]
For a probability density function, the total area under the graph is 1. This fixes the constant \(k\).
Step 2: Find k:
\[ \int_0^1 kx\,dx + \int_1^2 k\,dx + \int_2^3 (-kx + 3k)\,dx = 1 \]
First piece: \(\frac k2\). Second piece: \(k\). Third piece: \(k\left[-\frac{x^2}{2} + 3x\right]_2^3 = k\left[\left(-\frac92 + 9\right) - \left(-2 + 6\right)\right] = k\left(\frac92 - 4\right) = \frac k2\).
Total: \(\frac k2 + k + \frac k2 = 2k = 1\), so \(k = \frac12\).
Step 3: Required probability:
\[ P(2 < X \le 3) = \int_2^3(-kx + 3k)\,dx = \frac k2 = \frac14 \]
Step 4: Why the other options are wrong.
\(\frac12\) is the value of \(k\), not the probability. \(\frac13\) and \(\frac15\) do not follow from the areas \(\frac14, \frac12, \frac14\).
Final Answer:
The probability is \(\frac14\), option (C). \[ \boxed{\frac{1}{4}} \]
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