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Probability and Statistics

GATE CSE 2025 Set 1 · Q58

NAT2 marksEasy

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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) = \begin{cases} Cx^2, & \text{for } 1 \le x \le 4 \\ 0, & \text{for } x < 1 \text{ or } x > 4 \end{cases} The probability that xx lies between 2 and 3, i.e., P(2≤x≤3)P(2 \le x \le 3) is __________. (rounded off to three decimal places)

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Solution

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Official answer: 0.302 (±0.002) (±0.002)

Probability and Statistics

To find the probability, we first determine the normalization constant CC such that the total probability integral equals 11:

∫−∞∞P(x) dx=∫14Cx2 dx=1\int_{-\infty}^{\infty} P(x)\,dx = \int_{1}^{4} C x^2\,dx = 1

Evaluating the integral:

[Cx33]14=C3(43−13)=C3(64−1)=21C=1  ⟹  C=121\left[ C \frac{x^3}{3} \right]_{1}^{4} = \frac{C}{3} \left( 4^3 - 1^3 \right) = \frac{C}{3} (64 - 1) = 21C = 1 \implies C = \frac{1}{21}

Now, calculate the probability P(2≤x≤3)P(2 \le x \le 3):

P(2≤x≤3)=∫23Cx2 dx=[121⋅x33]23=163(33−23)=27−863=1963P(2 \le x \le 3) = \int_{2}^{3} Cx^2\,dx = \left[ \frac{1}{21} \cdot \frac{x^3}{3} \right]_{2}^{3} = \frac{1}{63} \left( 3^3 - 2^3 \right) = \frac{27 - 8}{63} = \frac{19}{63}
1963≈0.301587...≈0.302\frac{19}{63} \approx 0.301587... \approx 0.302

Rounded off to three decimal places, the value is 0.302 (range: 0.300 to 0.302).

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This Probability and Statistics previous year question is Q58 of the GATE CSE 2025 Set 1 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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Probability and Statistics has 46 GATE CSE PYQs across 23 papers from 2011 to 2026. 15 of them are 2-mark NAT questions like this one, and GATE CSE 2025 Set 1 has 3.

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Consider a probability distribution given by the density function P(x). P(x) = Cx2, & for 1 ≤ x ≤ 4 0, & for x 4 The probability that x lies between 2 and 3, i.e., P(2 ≤ x ≤ 3) is … (rounded off to three decimal places)

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