A stick of length \(1\) is broken into two pieces by cutting at a randomly chosen point. What is the expected length of the smaller piece?

- \(1 /8\)
- \(1 /4\)
- \(1 /e\)
- \(1 /\pi\)

TIFR 2019 GS Part A, Problem 20

Statistics

Moderate

Do you really need a hint? Try it first!

Can you use uniform probability distribution in this question?

Let’s let \(x\) denote the length of the smaller piece.

Then \(x\) is uniform on \([0, 1/2]\), as the split-point \(s\) is uniform on \([0, 1]\), but for split points past \(1/2\), the “smaller piece” \(x\) becomes \(1-s\) instead of \(s\). Try to calculate the expectation now.

Then \(x\) is uniform on \([0, 1/2]\), as the split-point \(s\) is uniform on \([0, 1]\), but for split points past \(1/2\), the “smaller piece” \(x\) becomes \(1-s\) instead of \(s\). Try to calculate the expectation now.

In this case, it’s continuous uniform prabability distribution, and we use

\(E(x) = \int x \cdot p(x) ~dx\)

where \(p(x)\) is the probability density

\(E(x) = \int x \cdot p(x) ~dx\)

where \(p(x)\) is the probability density

The probability density \(p(x)\), defined on \([0, 0.5]\), is given by

\(p(x) = 2\)

(its integral over the interval is exactly \(1\)). So

you need to compute

\(\int_0^{0.5} x \cdot 2 dx.\)

\(p(x) = 2\)

(its integral over the interval is exactly \(1\)). So

you need to compute

\(\int_0^{0.5} x \cdot 2 dx.\)

So the answer you will get after completing the integral is \(\frac 14\)

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