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Probability that a continuous random variable lies in an interval, from its p.d.f.

This type of question, asked 5 times: 14 May 2024 S1, 22 Apr 2025 S1, 20 Apr 2026 S1, 13 May 2026 S2, 14 May 2026 S1. All 5 questions below, oldest first.

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  1. Question 1

    14 May 2024 S1

    109
    If f(x)={3(1−2x2)4;0<x<10;otherwisef(x)=\begin{cases}\dfrac{3(1-2x^2)}{4}; & 0<x<1\\0; & \text{otherwise}\end{cases} is a probability density function of XX, then P ⁣(14<x<13)P\!\left(\dfrac{1}{4}<x<\dfrac{1}{3}\right) is:

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  2. Question 2

    22 Apr 2025 S1

    122
    The p.d.f. of a continuous random variable X is f(x)={x218 if −3<x<30 otherwise f(x) =\left\{ \begin{matrix} \frac{x^{2}}{18} & \text{~if~}- 3 <x< 3 \\ 0 & \text{~otherwise~} \end{matrix} \right. Then P[∣X∣<2]=P\lbrack|X| < 2\rbrack =

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  3. Question 3

    20 Apr 2026 S1

    99
    Let X be a continuous random variable with the probability density function(p.d.f.) given by f(x)={kx,0≤x<1k,1≤x<2−kx+3k,2≤x<30,otherwisef(x) = \begin{cases} kx, & 0 \le x < 1 \\ k, & 1 \le x < 2 \\ -kx + 3k, & 2 \le x < 3 \\ 0, & \text{otherwise} \end{cases} P(2<X≤3)=⋯P(2 < X \le 3) = \cdots

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  4. Question 4

    13 May 2026 S2

    99
    Let XX be a continuous r. v. with p. d. f. f(x)={kx3,0≤x≤10,otherwisef(x) = \begin{cases} kx^3, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases} then P(X≤34∣X>14)=…P\left(X \le \dfrac{3}{4} \mid X > \dfrac{1}{4}\right) = \ldots

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  5. Question 5

    14 May 2026 S1

    99
    Let X be a random variable with the probability density function defined by f(x)=1x2,1<x<∞f(x) = \dfrac{1}{x^2}, 1 < x < \infty =0= 0 otherwise. If C1={x:1<x<2}C_1 = \{x : 1 < x < 2\} and C2={x:4<x<5}C_2 = \{x : 4 < x < 5\} then P(C1∪C2)=P(C_1 \cup C_2) =

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