Question
Download Solution PDFLet X have the pdf \(\rm f(x)=\left\{\begin{matrix}1-|1-x|, 0<x<2& P\left(\frac{1}{3}\le X\le \frac{4}{3}\right)\\\ 0,& \rm otherwise\end{matrix}\right.\) is equal to:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFTo find the probability \(P ({1 \over 3} \leq x \leq {4 \over 3})\) using the given probability density function (pdf), we need to integrate the pdf over the specified range.
The pdf \(f(x) = 1 - |1 - x|\) is a piecewise function. Let's split the integration into two parts:
For \(0 < x < 1\):
\(f(x) = 1 - (1 - x) = x\)
For \(1 \leq x < 2\) :
\(f(x) = 1 - |1 - x| = 1 - (x - 1) = (2 - x)\)
Now, we can evaluate the probability: \(P ({1 \over 3} \leq x \leq {4 \over 3}) = \int\limits_{1\over 3}^{4\over3} f(x) dx \)
Evaluating the integrals: \([{x^2\over2}]_{1\over3}^{1} + [2x - {x^2\over 2}]_{1}^{4\over3}\)
Simplifying this, we have: \(({1\over 2} - {1\over 18}) + ({8 \over 3} - {8\over 9} - 2 + 1) = {13\over 18}\)
Hence, the required probability is \(13\over 18\).
Last updated on Jul 12, 2025
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