Probability Problem

Three numbers are chosen uniformly and independently at random from the interval [0,1] and arranged to form an ordered triple (a,b,c) with a<=b<=c. What is the probability that 4a+3b+2c<=1?

I came across this problem while going through some old math materials. Any ideas?

Note by Leandre Kiu
5 years, 1 month ago

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I hope this isn't an active Brilliant question. If so, please let me know.

We first consider the volume of the region of $$(x,y,z)$$-space for which $$0 \le x \le y \le z \le 1$$ and $$4x + 3y + 2z \le 1$$. This is easy to find: there are four inequalities, which define four planes that enclose a tetrahedron. In the $$yz$$-plane (i.e., $$x = 0$$), the given inequalities give the conditions $$y \ge 0$$, $$z \ge y$$, $$3y + 2z \le 1$$. This is a triangle whose vertices are $$(x,y,z) \in \{ (0,0,0), (0,\frac{1}{5}, \frac{1}{5}), (0,0,\frac{1}{2})$$. Thus its area is $$\frac{1}{20}$$. Now, the height of the tetrahedron as measured from this base is the maximum value of $$x$$ that satisfies the given constraints--this obviously occurs when $$x = y = z$$ and $$4x + 3y + 2z = 1$$, from which we immediately find $$x = \frac{1}{9}$$. Therefore, the total volume of the tetrahedron is $$\frac{1}{540}$$.

But this represents the probability that the three numbers chosen satisfy the given conditions without first being reordered in increasing sequence; i.e., that $$a = x$$, $$b = y$$, and $$c = z$$. Hence, the desired probability is $$3! = 6$$ times the volume found, since there are this many ways to permute three distinct elements--e.g., we could have $$a = y, b = z, c = x$$, or $$a = z, b = x, c = y$$, etc. Therefore, the desired probability is $$\frac{1}{90}$$.

- 5 years, 1 month ago