MHT CET · Maths · Probability
If \(\mathrm{f}(x)= \begin{cases}3\left(1-2 x^2\right) & ; 0 < x < 1 \ 0 & ; \text { otherwise }\end{cases}\) is a probability density function of \(\mathrm{X}\), then \(\mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}\right)\) is
- A \(\frac{75}{243}\)
- B \(\frac{23}{96}\)
- C \(\frac{179}{864}\)
- D \(\frac{52}{243}\)
Answer & Solution
Correct Answer
(C) \(\frac{179}{864}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}\right)= \int_{\frac{1}{4}}^{\frac{1}{3}} \mathrm{f}(x) \mathrm{d} x=\int_{\frac{1}{4}}^{\frac{1}{3}} 3\left(1-2 x^2\right) \mathrm{d} x \)
\( =\left[3 x-2 x^3\right]_{\frac{1}{4}}^{\frac{1}{3}} \)
\( =\left(1-\frac{2}{27}\right)-\left(\frac{3}{4}-\frac{1}{32}\right) \)
\( =\frac{1}{4}+\frac{1}{32}-\frac{2}{27} \)
\( =\frac{179}{864}\)
\( =\left[3 x-2 x^3\right]_{\frac{1}{4}}^{\frac{1}{3}} \)
\( =\left(1-\frac{2}{27}\right)-\left(\frac{3}{4}-\frac{1}{32}\right) \)
\( =\frac{1}{4}+\frac{1}{32}-\frac{2}{27} \)
\( =\frac{179}{864}\)
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