GATE Overflow Test Series | Probability | Test 1 | Question: 14

Question

A stick of length $2m$ is broken into two pieces, at a uniformly random chosen break point.The expected length of the larger piece (in meters) is _____

Comments

For more reading – https://math.stackexchange.com/questions/350679/we-break-a-unit-length-rod-into-two-pieces-at-a-uniformly-chosen-point-find-the/350863#350863

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Also read related comment by Dilip Sarwate from here.

Answer 1

When we break the stick into two pieces at a point $x,$ we will be left with one piece of length $x$ and a second piece of length $L - x.$  
Since their lengths add up to $L,$ whatever be the choice of $x,$ one piece will always be less than $ \frac{L}{2}$, and the other greater than $\frac{L}{2}.$

So, the larger piece will have a length ranging from $\frac{L}{2}$ to $L.$
If the break point $x$ is uniformly distributed throughout the length of the stick, then any point between $\frac{L}{2}$ and $L$ is equally likely and their mean gives $\frac{3L}{4}.$ So, the average length of the larger piece is $\frac{3L}{4} = 0.75 \times 2 = 1.5.$

Answer 2

Here uniform distribution is mentioned so answer by gatecse is better. But here is an alternative which can be used if some other distribution was mentioned:-

Expected length of larger part= length of part * probability of that length= xdx * $\frac{1}{2}$ 

xdx because probability of length being exactly x will be zero so we take length to be near x

probability of selecting any length is $\frac{1}{2}$ because of uniform distribution of length between 0 to 2.

 

Length of larger part= $\int_{0}^{1}(2-x)dx*\frac{1}{2} + \int_{1}^{2}xdx*\frac{1}{2} =\frac{3}{2}$

-->While integrating from length 0 to 1, length of larger part will be 2-x and from length 1 to 2 length of larger part will be x.

Answer 3

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