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Three random squares (Posted on 2021-06-23) Difficulty: 3 of 5
Given three random numbers a, b, and c
each a positive decimal fraction.

What is the probability that

a2+ b2+c2< 1 ?

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 2 of 3 |
These can be viewed as random points within a unit cube with one corner at the origin and three of its edges being segments of the x, y and z axes.

The question is equivalent to finding the probability that a point within this cube is within the corresponding octant of the unit sphere centered on the origin. This is equivalent to the probability (fraction of the points) within a 2x2 cube will lie within the sphere as a whole, as each octant has the same probability.

The volume of the unit (radius 1) sphere is 4*pi/3. The volume of the 2x2x2 cube is 8. Doing the division, the probability is pi/6 or about 0.523598775598299.

Simulation of a million trials:

clc, clearvars
succ=0;
for trial=1:1000000
   v=rand^2+rand^2+rand^2;
   if v<1 
       succ=succ+1;
   end
end
disp(succ/trial)

results in     0.523746

With 10 million trials  0.5235068



  Posted by Charlie on 2021-06-23 11:51:11
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