MHT CET · Maths · Probability
If the function \(f\) defined by \(f(x)=K\left(x-x^{2}\right)\) if \(0 < x < 1\)
\(=0 \quad, \quad\) other wise
is the p.d.f. of a r. v. \(X\), then the value of \(P\left(X < \frac{1}{2}\right)\) is
- A \(\frac{1}{4}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given \(f\) to a the p.df of a r.v.x.
\(\int f(x) d x=1 \Rightarrow \int_{0}^{1} K\left(x-x^{2}\right) d x=1 \)
\( K\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=1 \Rightarrow K\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow\) \(\frac{K}{6}=1 \)
\( K =6 \)
\( \therefore P\left(x < \frac{1}{2}\right)=\int_{0}^{\frac{1}{2}} 6\left(x-x^{2}\right) d x=\left[\frac{6 x^{2}}{2}\right]_{0}^{1}-\) \(\left[\frac{6 x^{3}}{3}\right]_{0}^{1} \)
\( =\left[3 x^{2}-2 x^{3}\right]_{0}^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\)
\(\int f(x) d x=1 \Rightarrow \int_{0}^{1} K\left(x-x^{2}\right) d x=1 \)
\( K\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=1 \Rightarrow K\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow\) \(\frac{K}{6}=1 \)
\( K =6 \)
\( \therefore P\left(x < \frac{1}{2}\right)=\int_{0}^{\frac{1}{2}} 6\left(x-x^{2}\right) d x=\left[\frac{6 x^{2}}{2}\right]_{0}^{1}-\) \(\left[\frac{6 x^{3}}{3}\right]_{0}^{1} \)
\( =\left[3 x^{2}-2 x^{3}\right]_{0}^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\)
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