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Probability and Statistics

GATE CSE 2025 Set 2 · Q65

NAT2 marksMedium

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The unit interval (0,1)(0,1) is divided at a point chosen uniformly distributed over (0,1)(0,1) in R\mathbb{R} into two disjoint subintervals. The expected length of the subinterval that contains 0.40.4 is ________. (rounded off to two decimal places)

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Official answer: 0.74

Probability and Statistics

Let X∼Uniform(0,1)X \sim \text{Uniform}(0, 1) be the random variable representing the division point in the interval (0,1)(0, 1).

The two disjoint subintervals created are (0,X)(0, X) and (X,1)(X, 1).

The point 0.40.4 belongs to:

  • (X,1)(X, 1) with length 1−X1 - X, if X≤0.4X \le 0.4.

  • (0,X)(0, X) with length XX, if X>0.4X > 0.4.

Let LL be the length of the subinterval containing 0.40.4:

L(X)={1−X,0<X≤0.4X,0.4<X<1L(X) = \begin{cases} 1 - X, & 0 < X \le 0.4 \\ X, & 0.4 < X < 1 \end{cases}

Since XX has a probability density function fX(x)=1f_X(x) = 1 for x∈(0,1)x \in (0, 1), the expected value E[L]E[L] is:

E[L]=∫00.4(1−x) dx+∫0.41x dxE[L] = \int_{0}^{0.4} (1 - x) \, dx + \int_{0.4}^{1} x \, dx

Evaluating the integrals:

∫00.4(1−x) dx=[x−x22]00.4=0.4−0.162=0.4−0.08=0.32\int_{0}^{0.4} (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_0^{0.4} = 0.4 - \frac{0.16}{2} = 0.4 - 0.08 = 0.32
∫0.41x dx=[x22]0.41=12−0.162=0.5−0.08=0.42\int_{0.4}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0.4}^{1} = \frac{1}{2} - \frac{0.16}{2} = 0.5 - 0.08 = 0.42

Summing the two parts:

E[L]=0.32+0.42=0.74E[L] = 0.32 + 0.42 = 0.74

Thus, the expected length rounded off to two decimal places is 0.74 (acceptable range: 0.70 to 0.80).

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This Probability and Statistics previous year question is Q65 of the GATE CSE 2025 Set 2 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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The unit interval (0,1) is divided at a point chosen uniformly distributed over (0,1) in R into two disjoint subintervals. The expected length of the subinterval that contains 0.4 is … (rounded off to two decimal places)

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