Randomly selected numbers on a continuous interval
Ekesh randomly chooses three numbers a , b , and c between 0 and 1 . If the probability P ( ∣ ∣ max ( a , b , c ) − min ( a , b , c ) ∣ ∣ ≥ 3 1 ) can be expressed as n m , where m and n are coprime positive integers, what is m + n ?
The answer is 47.
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1 solution
Good solution. It's different from my own.
In general, its trivial to prove that P ( m a x ( x 1 , x 2 , . . . , x n ) − m i n ( x 1 , x 2 , . . . , x n ) < p ) = p n + n ⋅ p ( n − 1 ) ⋅ ( 1 − p ) . Substituting p = 3 1 and n = 3 gives the answer, no stereometry required.
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@Ivo Zerkov - that's a good point, makes for an easier solution. I guess you can do something like done here (http://www.di.fc.ul.pt/~jpn/r/prob/range.html) to show that … but that doesn't seem trivial to me … why is it trivial ? thanks!
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If all n variables end up smaller than p (happens with probability p n ), the condition is certainly satisfied. Otherwise, the largest variable x k is larger than p . There’s n ways to pick x k , 1 − p probability this particular variable satisfies x k > p and p n − 1 probability the other n − 1 variables all end up in the interval ( x k − p , x k ) , which is what the condition equates to.
Adding up both possibilities leads to P ( m a x ( x 1 , x 2 , . . . , x n ) − m i n ( x 1 , x 2 , . . . , x n ) < p ) = p n + n ( 1 − p ) ( p n − 1 )
P ( max ( a , b , c ) − min ( a , b , c ) ≥ 1 / 3 ) = 1 − P ( max ( a , b , c ) ) − min ( a , b , c ) ≤ 1 / 3 ) = 1 − V o l u m e ( R ) , where R is the 3d region given by R = { ( a , b , c ) : max ( a , b , c ) ) − min ( a , b , c ) ≤ 1 / 3 , 0 ≤ a , b , c ≤ 1 } . After some thinking you realize that R is the region formed by translating the cube { ( a , b , c ) : 0 ≤ a , b , c ≤ 1 / 3 } from when its corner at ( 0 , 0 , 0 ) to when that corner is at ( 2 / 3 , 2 / 3 , 2 / 3 ) . The volume of this region is the size of this cube ( 1 / 3 ) 3 plus the volume of the "cylinder" formed by the translation. The cross-section of this cylinder is a regular hexagon with sidelength 2 / 2 7 ; this cross section is translated by 2 / 3 units. So the overall answer is
1 − ( 2 / 3 ∗ 3 ∗ 3 / 2 ∗ ( 2 / 2 7 ) 2 + ( 1 / 3 ) 3 ) = 2 0 / 2 7