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The question I have on hand is as follows :

We draw at random a number in interval [0,1] such that each number is "equally likely". Suppose we do the experiment two times (independently), giving us two numbers in [0,1]. What is the probability that the sum of these numbers is greater than 1/2?

My understanding is that since the experiments are independent, I am able to multiply the probability of each experiment with each other directly.

My attempt at this was to calculate probability of each experiment where x (the random number) is less than 1/4, thus giving me the probability of 1/16 when I multiply them. This would imply that there is a 15/16 chance my sum is greater than 1/2. However the answer is wrong since it is supposed to be 7/8.

Any help please? Thank you

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Choosing two numbers randomly (uniformly, independently) in the unit interval $[0,1]$ is the same as choosing a single point uniformly in the unit square $[0, 1]\times [0,1]$, and looking at its first and second coordinate.

Now, take a look at that square (you can even draw it, if you want). See if you can tell which points are such that their two coordinates add up to more than $\frac12$. (The line $x+y = \frac12$ is very relevant here, because it consists of the points where the two coordinates add up to exactly $\frac12$. That is what $x + y = \frac12$ really means, after all. You can draw that too to help you.) What's the area of that region?

As an extra exercise, you tried looking at the region where both the variables were smaller than $\frac14$. Can you draw that into the square? Can you see why your answer was wrong?

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    $\begingroup$ I get it now! With the square I drew, I would be splitting the square into 4 smaller squares and following that, cut the bottom left square at (0.5,0.5) and get 2 triangles, thus giving me 7/8. What I did initially was splitting the smaller square into 4 smaller parts, neglecting some parts of triangle I was supposed to get. Seems like a really handy method to deal with similar probability questions, thank you so much! $\endgroup$ – Wei Xiong Yeo Jun 14 '18 at 7:51
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    $\begingroup$ @WeiXiongYeo Excellent! Especially for uniform, independent variables (not necessarily with the same distribution) this technique works wonders. For other cases, it's a bit more tricky as the resulting point you choose doesn't have a uniform distribution on a rectangle. This is where integration (and measure theory) makes its appearance in probability theory, if you ever go that far. $\endgroup$ – Arthur Jun 14 '18 at 8:00

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