Aspire Faculty ID #15452 · Topic: MAH MCA CET 2025 (Shift 1) · Just now
MAH MCA CET 2025 (Shift 1)

The probability distribution function of a random variable $X$ is given by $f(x) = \dfrac{x}{18}, ; 0 \le x \le 6$ $= 0, \text{ otherwise}$ Then the value of $P(X > 2)$ is

Solution

$P(X > 2) = \int_{2}^{6} \dfrac{x}{18} , dx = \dfrac{1}{18} \left[\dfrac{x^2}{2}\right]_2^6 = \dfrac{1}{36}(36 - 4) = \dfrac{32}{36} = \dfrac{8}{9}$

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