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If the probability density function of a random variable $X$ is $f(x)=\frac{x}{2}$ in $0 \leq x \leq 2$, then $P(X>1.5 \mid X>1)$ is equal to
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The correct answer is:
$\frac{7}{16}$
$\int_{1.5}^{2} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}$, where $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}$
$\begin{array}{l}
\int_{1.5}^{2} \frac{x}{2} d x=\frac{1}{2} \cdot \int_{1.5}^{2} x d x=\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{1.5}^{2} \\
=\frac{1}{4}[4-2.25]=\frac{1}{4} \times[1.75]=\frac{175}{400}=\frac{7}{16}
\end{array}$
$\begin{array}{l}
\int_{1.5}^{2} \frac{x}{2} d x=\frac{1}{2} \cdot \int_{1.5}^{2} x d x=\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{1.5}^{2} \\
=\frac{1}{4}[4-2.25]=\frac{1}{4} \times[1.75]=\frac{175}{400}=\frac{7}{16}
\end{array}$
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