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Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ .
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The length of the shorter stick can be from $0$ to $0.5$ (because if it is greater than $0.5,$ it is no longer a shorter stick).

 This random variable $L$ (length of shorter stick) follows a uniform distribution, and hence probability density function of $L$ is $\dfrac{1}{0.5-0}= 2$ for all lengths in range $0$ to $0.5$

Now expected value of $L = \int_{0}^{0.5} L*p(L) dL = \int_{0}^{0.5} L*2 dL = 2*\left[\dfrac{L^2}{2}\right]^{0.5}_0 = 0.25$
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@Happy mittal... can you please explain how did you calculated probability density function...??
In all of the links you provided, stick/rope is broken into 3 pieces, but here in question, it is broken into 2 pieces.
@Vicky : pdf of a uniform distribution from a to b is 1/(b-a).
How could someone figure out  that it could be solved using Uniform Distribution unless there is tag !? Oops
question directly taken from sheldon ross
@sushmita See Cormen, Rosen, Ross, Dragon book, Korth etc.: many questions will be directly taken in GATE. Even same examples in text they use in GATE.
yes arjun sir. i wish i had time to see all those problems. But i lack time. Getting super nervous. :-(((((((.
@Arjun Sir, I also have same doubt as pC. How to know whether the question has to be solve by Uniform Distribution function or some else?,

Because question says so

at a point chosen uniformly at random

In GATE all information and hints will always be there in question. But many aspirants are used to solving substandard questions prepared in test series and hence skip those vital information.


For better intuition ..:)

Don't it should be 0 to 0.49 because they are asking about shorter stick length. but if we take 0 to 0.5 then it includes shorter as well as equal length?

What if the stick is broken randomly at two points? My approach:

To be shortest: length < 1/3

Pdf = 3 (using 1/b-a)

$3 \int_{0}^{1/3} L dx = 1/6$

Is 1/6 correct? Below link says it is 1/9:

@shubhanshu and if we consider .5 to be "short" then the limits should go from 0 to 1 making 0.5 the answer? is my intuition correct?
The length of the shorter stick will be in the interval $[0,0.5]$ with uniform distribution.

If it's greater than 0.5, it won't be a shorter stick.

Expected length of shorter stick=$\frac{0+0.5}{2}=0.25$

@Happy Mittal Can we also do this like this: First we calculate the expected value of shorter stick ( integration of x time 1 with limits 0 to 1. We now use this expected value as limits to a new expectation, which will the expected value of the shorter stick. I got the same answer using that. Just wanted to confirm.


The length of the shorter stick can be from 0 to 0.5 (because if it is greater than 0.5, it is no longer a shorter stick).

Length of the shorter stick can be from 0.5 to 1 as well. Then why we are considering only 0 to 0.5


Length of the shorter stick can be from 0.5 to 1 OR  0 to 0.5. so even if you take range 0.5 to 1 you will get same answer.

But in expectation we consider both the scenarios, so why here only one part is being considered.
Both scenarios are identical so expectation will not change.

if you consider scenario 1 then you get 0.25, if you consider scenario 2 then you get 0.25 .

expecation = (scenario1 + scenario 2 )/2 = (0.25 +0.25) /2 = 0.25

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