Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?

Also what is probability of sum of them being less than $1$?

I think the answer should be $\frac{1}{2}$, but I have no idea.

EDIT: I should mention that the question is about a general distribution, and the numbers selected are independent.

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    $\begingroup$ The question can't be answered in its present form because you haven't specified a probability distribution. Did you intend to imply that the numbers are independent and uniformly distributed? $\endgroup$ – joriki Jan 23 '13 at 21:35
  • $\begingroup$ @joriki: Actually I need a general solution, for any distribution, but you can assume they are independent. $\endgroup$ – PLuS Jan 23 '13 at 21:38
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    $\begingroup$ @PLuS: Then you should probably add that to the question. We get so many questions that implicitly assume uniformity and independence that people tend to automatically make that assumption, as you can see from the answers already given. $\endgroup$ – joriki Jan 23 '13 at 21:40
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    $\begingroup$ @amWhy: Why did you replace general distribution with uniform distribution?? You have changed the meaning of the question. Please be more careful with your edits! $\endgroup$ – Rahul Jan 23 '13 at 22:12
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    $\begingroup$ I meant general distribution and that was what I had written in my edit! $\endgroup$ – PLuS Jan 24 '13 at 22:47

For the general case, if $X$ and $Y$ are two independent, identically distributed variables with probability distribution function $f$ supported in $[0,1]$, then the probability that their sum exceeds $1$ is $$\begin{align} P[X+Y\ge1]=\iint\limits_{\substack{0\le x\le1\\0\le y\le1\\{x+y\ge1}}} f(x)f(y)\,\mathrm dx\,\mathrm dy=\int_0^1f(x)\left(\int_{1-x}^1f(y)\,\mathrm dy\right)\,\mathrm dx=\int_0^1f(x)\big(1-F(1-x)\big)\,\mathrm dx, \end{align}$$ where $F(t)=\int_0^tf(u)\,\mathrm du$ is the cumulative distribution function of the random variables. I don't think this simplifies any further.

(Sanity check: For the uniform distribution, $f$ is identically $1$ in $[0,1]$, so $F(t)=t$ and you get $P[X+Y\ge1] = \int_0^1\big(1-(1-x)\big)\,\mathrm dx=\int_0^1 x\,\mathrm dx=1/2$, as expected.)

  • $\begingroup$ Thanks, this was what I was looking for, is there a proof for that? $\endgroup$ – PLuS Jan 24 '13 at 22:44
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    $\begingroup$ @PLuS: What part do you feel requires a proof? The domain of the first integral is exactly the range of $x$ and $y$ whose sum is more than $1$. The domains of the second integrals come from $x+y\ge1\implies y\ge1-x$. $\endgroup$ – Rahul Jan 24 '13 at 22:53
  • $\begingroup$ @PLuS: A slight simplification is possible if you have both the PDF and the CDF handy; see my edit. $\endgroup$ – Rahul Feb 3 '13 at 20:46

Choosing two numbers in (0,1) is the same that choosing a point in the square $(0,1)^2$. Draw the half-plane $x+y\ge 1$ and you will see the answer...

enter image description here

  • $\begingroup$ That's a nice solution, is there another way of doing it without considering square? $\endgroup$ – PLuS Jan 23 '13 at 21:37
  • $\begingroup$ My answer below does not explicitly mention the square. (But certainly you should be aware of the square.) $\endgroup$ – Michael Hardy Jan 24 '13 at 0:19
  • $\begingroup$ @Michael Hardy: Sorry, I cant spare a square ;. $\endgroup$ – gary Mar 25 '15 at 7:32

If you mean uniformly distribued random numbers, then your question should say that. And if they're supposed to be independent, that should also be mentioned.

Supposing those assumptions to be the case, we have an argument from symmetry: Let $X$, $Y$ be the two uniformly distributed random variables. Let $U=1-X$ and $V=1-Y$. Then $U$, $V$ are also uniformly distributed in $(0,1)$ and are independent. So $P(U+V>1)$ is the same as $P(X+Y>1)$. But $U+V>1$ if and only if $X+Y<1$. So $P(X+Y>1)=P(U+V>1)=P(X+Y<1)$. If the probabilities of two complementary events are equal, the each probability is $1/2$.

(We need not worry about the event that $X+Y$ is exactly $1$ since the probability of that is $0$.)


To pursue manu-fatto's wonderful suggestion/answer, and given your clarification that the numbers are drawn randomly and independently, (but nonetheless assuming that we are dealing with a uniform distribution):

(1) It amounts to the probability of a point $(x, y)$ in the unit square lying in the shaded region (with respect to the unit square): $\iff x + y > 1\quad x \in (0, 1), y \in (0, 1)$

enter link description here

i.e. probability = $\dfrac{1}{2}$ the area of the unit square $= \dfrac{1}{2}$

(2) What amounts to the probability of a point $(x, y)$ in the unit square lying in the shaded region $\iff x + y \lt 1 \quad x\in (0, 1), y\in (0, 1)$

enter image description here

i.e. probability = $\dfrac{1}{2}$ the area of the unit square $= \dfrac{1}{2}$

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    $\begingroup$ I see, you changed the question so that your answer would make sense. -1. $\endgroup$ – Rahul Jan 23 '13 at 22:20
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    $\begingroup$ That's not what I did, @Rahul. My "answer" is only a comment, but too long to post as a comment. $\endgroup$ – Namaste Jan 23 '13 at 22:22
  • $\begingroup$ PLuS: click on the link to "uniform distribution" in this answer to see what we mean by making explicit this qualification in many of the answers. I'm not clear what you mean by "general distribution" in your edit; I assumed that you used "general" to mean "uniform", but you should edit your post if that is what you mean, once you read what is meant by a uniform distribution. $\endgroup$ – Namaste Jan 23 '13 at 22:38
  • $\begingroup$ @amWhy: Which book contains this type of problems ? can you recommend any? $\endgroup$ – Chinnapparaj R Dec 22 '18 at 11:30

There should not an answer to this question. Choosing randomly two numbers does not tell us any probability distribution. Maybe you meant choose randomly, uniformly and independently, as the edit says.

Only given that it is uniform distribution, their pdfs:

$f_{X} (t) = 0$ if t outside (0,1)

$f_{X} (t) = 1$ if $0\le t \le 1$

$f_{Y} (t) = 0$ if t outside (0,1)

$f_{Y} (t) = 1$ if $0\le t \le 1$

Introduce new random variable Z = X + Y, pdf of Z is convolution of X and Y

$f_{Z}(t) = \int_{-\infty}^\infty f_X(w)f_Y(t-w)dw = \int_{0}^t f_X(w)f_Y(t-w)dw$

Here discuss t

$t<0$: $f_Z(t) = 0$

$0\le t\le1$: $f_Z(t) = t/2$

$1\le t\le2$: $f_Z(t) = 1-t/2$

$t>2$: $f_Z(t) = 0$

Now to get what you want, integrate:

Probability of sum more than 1

$\int_1^\infty f_Z(t)dt = \int_1^2 f_Z(t)dt = 1/2$

Probability of sum less than 1

$\int_{-\infty}^1 f_Z(t)dt = \int_0^1 f_Z(t)dt = 1/2$ or simply use 1-1/2 from last result.


The mean value of $a+b$ if $a$ and $b$ are the random numbers would be $\frac{1}{2} + \frac{1}{2} = 1$, and it will be symmetric around the mean, so yes, the probability is $\frac {1}{2}$.


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