What is the expected length of the smaller piece? 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?
(a)$1 /8$
(b)$1 /4$
(c)$1 /e$
(d)$1 /π$
This is a question from an exam in India. I am bit confused to see this question and not getting any clue. What I know is $E(x)=\sum xp(x)$. Now how to use that theory here?
 A: Given that the length of the stick is $1$. We break the stick into two pieces at some point $x$. Then we will be left with one large piece and one small piece.
Let the length of one piece of stick be $x$, then the other piece will be $1-x$. When we add up, we get length $1$. So far so good.
Note that one piece will always be less than $\dfrac12$ and the other piece will be greater that $\dfrac12$. So, the smaller piece will have length ranging from $0$ to $\dfrac12$.
If you break point $x$ is uniformly distributed all through the length, then we are taking the average of $0$ and $\dfrac12$ which tells us the the expected length of the smaller piece of the stick is $\dfrac14$
A: 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 $s$. 
The formula you've got is great when the probability space is discrete; in this case, it's continuous, and we use
$$
E(x) = \int x \cdot d(x) ~dx
$$
where $d$ is the probability density rather than the probability mass. 
The probability density $d$, defined on $[0, 0.5]$, is given by 
$$
d(x) = 2
$$
(its integral over the interval is exactly $1$). So
you need to compute
$$
\int_0^{0.5} x \cdot 2 dx.
$$
A: It looks like a question to be quickly answered within a multiple-choice questionnaire.
So, an "optical" solution without any further calculation or reasoning might be helpful, as well:

