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?

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TopNewestI 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 \( \frac{1}{90} \).

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