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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
Let $\mathrm{f}(x)=\left\{\begin{array}{l}\frac{1}{x^2}, 1<x<\infty \\ 0, \text { otherwise }\end{array}\right.$ be the p.d.f. of a r.v. X. If $\mathrm{C}_1=\{x: 1<x<2\}$ and $\mathrm{C}_2=\{x: 4<x<5\}$, then $\mathrm{P}\left(\mathrm{C}_1 \cup \mathrm{C}_2\right)=$
  • A
    $\frac{1}{20}$
  • B
    $\frac{7}{20}$
  • ✓
    $\frac{11}{20}$
  • D
    $\frac{13}{20}$
✓

Answer

Correct option: C.
$\frac{11}{20}$
(C)
$\mathrm{P}\left(\mathrm{C}_1 \cup \mathrm{C}_2\right)$
$ =\mathrm{P}\left(\mathrm{C}_1\right)+\mathrm{P}\left(\mathrm{C}_2\right)$
$=\int_1^2 \mathrm{f}(x) \mathrm{d} x+\int_4^5 \mathrm{f}(x) \mathrm{d} x$
$=\int_1^2 \frac{1}{x^2} \mathrm{~d} x+\int_4^5 \frac{1}{x^2} \mathrm{~d} x$
$ =\left[\frac{-1}{x}\right]_1^2+\left[\frac{-1}{x}\right]_4^5$
$=-\frac{1}{2}+1-\frac{1}{5}+\frac{1}{4}$
$ =\frac{11}{20}$

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