Q99 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics

Given the probability density function (p.d.f.) of the random variable X, $f(x) = \dfrac{1}{2a}$, $0 < x < 2a$, $a > 0$
$= 0$, otherwise, then which of the following is correct ?

  • A
    $P\left(X < \dfrac{a}{2}\right) = P\left(X > \dfrac{a}{2}\right)$
  • B
    $P\left(X < \dfrac{a}{2}\right) < P\left(X > \dfrac{3a}{2}\right)$
  • C
    $P\left(X < \dfrac{a}{2}\right) > P\left(X > \dfrac{3a}{2}\right)$
  • Correct answer
    D
    $P\left(X < \dfrac{a}{2}\right) = P\left(X > \dfrac{3a}{2}\right)$

Solution

The given probability density function is $f(x) = \dfrac{1}{2a}$ for $0 < x < 2a$, and $0$ otherwise.

We calculate the probability $P\left(X < \dfrac{a}{2}\right)$:

$P\left(X < \dfrac{a}{2}\right) = \int_{0}^{\dfrac{a}{2}} f(x) dx = \int_{0}^{\dfrac{a}{2}} \dfrac{1}{2a} dx = \dfrac{1}{2a} [x]_{0}^{\dfrac{a}{2}} = \dfrac{1}{2a} \left(\dfrac{a}{2} - 0\right) = \dfrac{1}{4}$

Next, we calculate the probability $P\left(X > \dfrac{3a}{2}\right)$:

$P\left(X > \dfrac{3a}{2}\right) = \int_{\dfrac{3a}{2}}^{2a} f(x) dx = \int_{\dfrac{3a}{2}}^{2a} \dfrac{1}{2a} dx = \dfrac{1}{2a} [x]_{\dfrac{3a}{2}}^{2a} = \dfrac{1}{2a} \left(2a - \dfrac{3a}{2}\right) = \dfrac{1}{2a} \left(\dfrac{a}{2}\right) = \dfrac{1}{4}$

Comparing the two probabilities, we get:

$P\left(X < \dfrac{a}{2}\right) = P\left(X > \dfrac{3a}{2}\right) = \dfrac{1}{4}$

Answer: $P\left(X < \dfrac{a}{2}\right) = P\left(X > \dfrac{3a}{2}\right)$

Hence, the correct answer is option: 'D'.

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