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Official PYQ · 2025Set 1 · Q48Probability: Random Variables and DistributionsNAT2 marks

GATE CS 2025 Set 1 Q48 — Probability: Random Variables and Distributions

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Consider a probability distribution given by the density function P(x)P(x).
P(x)={Cx2, for 1≤x≤40, for x<1 or x>4P(x)=\left\{\begin{aligned} C x^{2}, & \text{ for } 1 \leq x \leq 4 \\ 0, & \text{ for } x<1 \text{ or } x>4 \end{aligned}\right.
The probability that xx lies between 22 and 33, i.e., P(2≤x≤3)P(2 \leq x \leq 3) is _________. (rounded off to three decimal places)
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Answer: 0.300:0.302

To find the probability P(2≤x≤3)P(2 \leq x \leq 3), we first determine the normalization constant CC using the property that the total area under a probability density function is 11:

∫14P(x) dx=1  ⟹  ∫14Cx2 dx=1\int_{1}^{4} P(x) \, dx = 1 \implies \int_{1}^{4} C x^2 \, dx = 1

Evaluating the integral:

C[x33]14=1  ⟹  C(64−13)=1  ⟹  21C=1  ⟹  C=121C \left[ \frac{x^3}{3} \right]_{1}^{4} = 1 \implies C \left( \frac{64 - 1}{3} \right) = 1 \implies 21C = 1 \implies C = \frac{1}{21}

Now, we calculate the probability that xx lies between 22 and 33:

P(2≤x≤3)=∫23121x2 dx=121[x33]23P(2 \leq x \leq 3) = \int_{2}^{3} \frac{1}{21} x^2 \, dx = \frac{1}{21} \left[ \frac{x^3}{3} \right]_{2}^{3} =163(33−23)=1963≈0.301587= \frac{1}{63} (3^3 - 2^3) = \frac{19}{63} \approx 0.301587

Rounding off to three decimal places, we get 0.302 (the acceptable range is 0.3000.300 to 0.3020.302).

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