Probability that the sum of 'n' positive numbers less than 2 is less than 2 I'm a high school student,
And I stumbled across this problem
Q) if you arbitrarily choose 3 real positive numbers less than or equal to 2
What is the probability that their sum is less than or equal to 2?
In my course I have learnt to solve 2 number based probability problems by using the ratio of the areas plotted by their graphs.
This was an application question that required me to plot it in 3 dimensions and find the ratio of their volumes to get the required probability, and so I managed to pull it off and got the required answer.
My question then arose... How do I deal with N dimensions? As I only have the knowledge to visualize 3 dimensions at my level of Math,
Can anyone help me out?
I am really curious to see
1) the actual function and how it's growth is
2) the way you solve such kind of problems! (some kind of multivariable calculus??)
Thanks! 
 A: This of course depends on how you choose the numbers.
We could assume $X_1$, $X_2$, $X_3$ are independent random variables
each uniformly distributed on the interval $[0,2]$.
I'd rather consider $Y_1=X_1/2$ etc., and ask for the probability
that $Y_1+Y_2+Y_3<1$. Then $(Y_1,Y_2,Y_3)$ is a point uniformly
distributed in the unit cube $C$
(with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ etc.)
Of course $C$ has volume $1$, and the probability you seek is
the volume of the tetrahedron $T$ with vertices
$(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$.
Using the formula that the volume of a tetrahedron is a third
of the area of the base times the height, then this is $1/6$.
In $n$ dimensions the probability is the $n$-volume of an
$n$-simplex with vertices $(0,0,\ldots,0)$ and the $n$ standard
unit vectors. You can compute its volume when you know that
the volume of an $n$-simplex is $1/n$ times the $(n-1)$-volume
of a face times the corresponding height.
A: I'm not a math guy or a statistician, but I'm an engineer. In engineering, often we don't care about the exact result achieved by integrating 4 times, but rather a way to just get close enough. Here is what I did to get close enough using python.
import random
# You can try changing this number.
n = 5

#a large number of trials. i.e. how many times we are going to 'roll the dice'?
trials = 500000 

successes = 0 # number of times the sum of n random numbers between 0 and 2 are less than 2
failures = 0 # number of times the sum of n random numbers between 0 and 2 are NOT less than 2

for i in range(0, trials): 
    #sum up n random numbers
    total = sum([random.uniform(0,2) for j in range(0,n)]) 

    # if the sum of our numbers is less than or equal to two, let's call it a success, otherwise a failure
    if total <= 2: successes += 1 
    else: failures += 1

# Finally, divide our number of successes by the total
probability = successes / (successes + failures) 
print("Approx. probability that {} random numbers between 0 and 2 sum up to be less than 2:".format(n), probability)

Output:
n = 2: 0.501466
n = 3: 0.167362
n = 4: 0.042046
n = 5: 0.008338
n = 6: 0.001262
n = 7: 0.000192
n = 8: 2.8e-05

Basically, the code above simulates 500,000 trials of the experiment and keeps track of how many successes and failures there are then divides the number of successes by the number of trials to estimate a probability.
