Six variables x i are chosen uniformly at random in the interval [ 0 , 1 ] .
Find the probability that x 1 + x 2 + x 3 + x 4 + x 5 + x 6 < 1 . If the answer is of the form b a , where a and b are coprime positive integers, what is a + b ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Solving this is equivalent to finding the volume of the unit 5-simplex in 6 dimensions (essentially, a "right tetrahedron" extrapolated into 6 dimensions).
A unit 2-simplex is 3 dimensions is shown below.
This is equivalent to finding the geometric probability via the volume of the 6-dimensional tetrahedron :)
Log in to reply
Ah yes! :0)
(But not a regular tetrahedron, right? More of a "right" tetrahedron)
Log in to reply
Yup. The heights of 1 tells us that the volume of the tetrahedron (calculated itertively over the dimensions) is 1 × 2 1 × 3 1 × 4 1 × 5 1 × 6 1 .
Right, this scenario in particular is the standard n-simplex.
While the multiple integration approach works, the formulae as stated are a little more complicated unless you define u-z a little more clearly.
Is there a way of solving this using characteristic functions? As a rule they make these sorts of problems much easier to resolve - though I have yet to do that in this case
Log in to reply
Good point... I should probably update the solution to be a bit more clear.
Log in to reply
There's probably a substitution that turns this into the six-time integration of 1. Integrate (A-x)^n between 0 and A still gives (1/n)*A^(n+1). And if A is successively x1, x1+x2, x1+x2+x3,.. then we get the same result
is there uhhhh an easier way to do this without calculus
I think the proposed solution is the answer of; the probability of chosing numbers obtaining the given start up condition, in order. If you choose 6 different numbers between 0 and 1; all chosen set will obtain that condition. It doesn't need to be in order to obtain the given condition. The problem is not; what's the probability of chosing the numbers from a bag with exact order?
What is the probability of total sum being between 3 and 4? Is it same as being smaller than 1?
Log in to reply
Interesting follow up question... I believe the probability of being between 3 and 4 will be much higher!
"In general, we can solve this by nested integrations, where you integrate over all the possibilities for the first value being in this range. Then integrating over all the values where the second value is in this range (less the first value), then integrating over the third value being in this range (less the first two values) etc. until we have integrated over all the variables. The value of that integral will represent the final probability." I didn't understand why this would give the final probability. We want the sum of the 6 variables to be less than 1. How is that condition is equivalent to the above condition? Thanks.
Log in to reply
I agree with you. Apologies for not formatting, since I'm not sure how to on mobile, but it seems to me that you just calculated the probability that u <v <w <x <y <z. So it seems like 1/6! Should be the denominator ofthe true probability, while the numerator is the probability that the sum is less than 1 given that the variables are ordered by magnitude.
Log in to reply
I'm not sure why you think that u < v < w < x < y < z. How about if u is 0.9 and the rest are 0.01?
Log in to reply
@Geoff Pilling – Sorry, I got the inequality backwards, it's actually u >v>w>x>y>z. However, the issue here is that the case u = 0.9, v = 0.8, and the rest 0 is included in the integral above, while that case clearly doesn't satisfy the conditions of the problem since the sum is greater than 1. Or to dramatize the problem further, u=v=w=x=y=z=1 is also part of the domain of the integral, but those numbers sum up to 6, which is quite obviously larger than 1. Again, the integral above is the probability that u>v>w>x>y>z, and you need to also find the probability given u>v>w>x>y>z that their sum is less than 1 in order to solve this problem
Log in to reply
@Kyle Coughlin – Yes. You are correct that I need to clarify the solution... Will work on that.
The sum of n independent random variables with uniform distribution is the Irwin-Hall distribution, whose cumulative distribution function is: 6 ! 1 ∗ ( ( − 1 ) 0 ∗ ( 6 C 0 ) ∗ ( 1 − 0 ) 6 + ( − 1 ) 1 ∗ ( 6 C 1 ) ∗ ( 1 − 1 ) 6 )
Let X be the sum of the 6 random variables, then: P(X<1)=P(X<1)= 7 2 0 1
a+b=720+1= 721
Not a creative solution thought :(
Oh, so a triangular distribution is a special case of this distribution? Thanks, I learned something new!
let us consider no line in 0 to 1 is made up of n points (mark them 1 to n) choosing any 6 pts we have way n^6 choice now let us find no of ways x1+x2+x3+x4+x5+x6<n(marked values) we get it as (n-1)C6 so divide and find limit n tends to infinity. :P
does it equal to the probability 5<x+y+z+u+v+w<6 ?
I don't understand what you're saying.
now let us find no of ways x1+x2+x3+x4+x5+x6<n(marked values)
What does this mean? What marked values are you referring to?
Limit n tends to infinity? Why is this needed?
Log in to reply
I mean we have divided number line from 0 to 1 into n pts and mark them 0 to n with pt at x=1 marked as n so what we have to find no of ways any 6 pts can be selected=n^6 now since its provided that x1+x2++x3+x4+x5+x6<1 but (think using length) so marked value of x1 be its length and so so length of x1+x2++x3+x4+x5+x6< length of 1 ie x1+x2+x3+x4+x5+x6<n so we get (n-1)C6 ways P=(n-1)C6 /n^6 but since no line consists of infinite pt so n---->infinity we get P=1/61
Log in to reply
Ah got it got it thanks
Log in to reply
@Pi Han Goh – Using the same method got:
for picking two numbers n → ∞ lim n 1 ∑ i = 1 n − 1 ( n − i ) = 2 1
for picking three numbers got n → ∞ lim n 3 ∑ i = 1 n − 2 ∑ j = 1 n − i − 1 ( n − ( i + j ) )
do you know how to generalize it for picking k numbers? or prove it equals to k ! 1 ? I got stuck there..
@Pi Han Goh – pls upvote if u liked it!!
this is the best solution for this question
Solving for all n (here n = 6 ) with S n = ∑ i = 1 n x i :
∫ ∀ i ∈ [ 0 , n ] 0 < x i < 1 , S n < 1 d x 1 . . . d x n = ∫ ∀ i ∈ [ 0 , n − 1 ] 0 < x i < 1 , 0 < x n < 1 − S n − 1 , S n − 1 < 1 d x 1 . . . d x n = ∫ ∀ i ∈ [ 0 , n − 1 ] 0 < x i < 1 , S n − 1 < 1 ( 1 − S n − 1 ) 1 d x 1 . . . d x n − 1 = ∫ ∀ i ∈ [ 0 , n − 2 ] 0 < x i < 1 , S n − 2 < 1 2 ( 1 − S n − 1 ) 2 d x 1 . . . d x n − 2 = . . . = n ! 1
So the answer is n ! + 1 , here 7 2 1 .
The sum of the variables on which you integrate. The trick is that you integrate them one by one, so Sn, Sn-1 etc. are constant when integrating for xn+1, xn etc.
Log in to reply
Ah got it. I'ts surprising to see that it simplifies to such a simple expression, when the integral looks so complicated at first glance
I don't understand what you're saying. What does S n = ∑ i = 1 n x i represent?
Problem Loading...
Note Loading...
Set Loading...
Consider the closed interval [0,1]. In general, we can solve this by nested integrations, where you integrate over all the possibilities for the first value being in this range. Then integrating over all the values where the second value is in this range (less the first value), then integrating over the third value being in this range (less the first two values) etc. until we have integrated over all the variables. The value of that integral will represent the final probability.
For example for six variables, x i and u , v , w , x , y and z defined as this:
we would have something like:
P = ∫ 0 1 ∫ 0 u ∫ 0 v ∫ 0 w ∫ 0 x ∫ 0 y d z d y d x d w d v d u
P = ∫ 0 1 ∫ 0 u ∫ 0 v ∫ 0 w ∫ 0 x y d y d x d w d v d u
P = ∫ 0 1 ∫ 0 u ∫ 0 v ∫ 0 w 2 x 2 d x d w d v d u
P = ∫ 0 1 ∫ 0 u ∫ 0 v 6 w 3 d w d v d u
P = ∫ 0 1 ∫ 0 u 2 4 v 4 d v d u
P = ∫ 0 1 1 2 0 u 5 d u
P = 7 2 0 1
P = 6 ! 1
1 + 7 2 0 = 7 2 1