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GATE CS 2025 Set 2 Q55 — Probability: Random Variables and Distributions

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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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Answer: 0.70:0.80

Let the division point be represented by the random variable XX, which is uniformly distributed over the interval (0,1)(0, 1). The probability density function of XX is:

f(x)=1for x∈(0,1)f(x) = 1 \quad \text{for } x \in (0, 1)

The point XX divides the interval (0,1)(0, 1) into two subintervals: (0,X)(0, X) and (X,1)(X, 1).

The subinterval containing 0.40.4 depends on the position of XX:

  • If X≤0.4X \le 0.4, the subinterval containing 0.40.4 is (X,1)(X, 1), which has a length of 1−X1 - X.
  • If X>0.4X > 0.4, the subinterval containing 0.40.4 is (0,X)(0, X), which has a length of XX.

Thus, the length of the subinterval containing 0.40.4, denoted by L(X)L(X), is defined piecewise as:

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

The expected length E[L(X)]\mathbb{E}[L(X)] is calculated as follows:

E[L(X)]=∫01L(x)f(x) dx=∫00.4(1−x) dx+∫0.41x dx\mathbb{E}[L(X)] = \int_{0}^{1} L(x) f(x) \, dx = \int_{0}^{0.4} (1 - x) \, dx + \int_{0.4}^{1} x \, dx

Evaluating the two integrals separately:

∫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

Adding the two parts together:

E[L(X)]=0.32+0.42=0.74\mathbb{E}[L(X)] = 0.32 + 0.42 = 0.74

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