MHT CET · Maths · Probability
The p. d. f. of a continuous random variable \(\mathrm{X}\) is given by
\(f(x)=\frac{1}{2} \quad\) if \(\quad 0 < x < 2\)
\(=0 \quad\) otherwise
and if \(a=P\left(X < \frac{1}{2}\right), b=P\left(X>\frac{3}{2}\right)\), then relation between \(a\) and \(b\) is
- A \(a-b=0\)
- B \(2a-b=0\)
- C \(3a-b=0\)
- D \(a-2 b=0\)
Answer & Solution
Correct Answer
(A) \(a-b=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} a &=PX < \frac{1}{2}=\int f(x) d x \\ &=\int f(x) d x-\int f(x) d x=0+\int \frac{1}{2} d x \\ &=\frac{1}{2}[x] \frac{1}{2}=\frac{1}{2} \frac{1}{2}-0 \quad=\frac{1}{4}=a=\frac{1}{4} \\ b &=P\left(X>\frac{3 a}{2}=\int f(x) d x\right.\\ &=\int_{\frac{3}{2}}^{2} f(x) d x+\int_{2}^{\infty} f(x) d x=\int_{\frac{3}{2}}^{2} \frac{1}{2} d x+0 \\ &=\frac{1}{2}[x]_{\frac{3}{2}}^{2}=\frac{1}{2}\left(2-\frac{3}{2}\right)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \Rightarrow b=\frac{1}{4} \\ \therefore & a-b=\frac{1}{4}-\frac{1}{4}=0 \quad \end{aligned}\)
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