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

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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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.)

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…

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): $\iff x + y > 1\quad x \in (0, 1), y \in (0, 1)$

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 $\iff x + y \lt 1 \quad x\in (0, 1), y\in (0, 1)$

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

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 $\frac {1}{2}$.

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.