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# probability problem

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The numbers \$x_1,\$ \$x_2,\$ \$x_3,\$ \$x_4\$ are chosen at random in the interval \$[0,1].\$  Let \$I\$ be the interval between \$x_1\$ and \$x_2,\$ and let \$J\$ be the interval between \$x_3\$ and \$x_4.\$  Find the probability that intervals \$I\$ and \$J\$ both have length greater than \$3/4\$.

Oct 9, 2023

#1
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To find the probability that intervals I and J both have length greater than 3/4, we can use the following steps:

Calculate the probability that interval I has length greater than 3/4. Since x1​ and x2​ are chosen independently at random in the interval [0,1], the length of interval I is uniformly distributed between 0 and 1. Therefore, the probability that interval I has length greater than 3/4 is 1−01−3/4​=1/4​.

Calculate the probability that interval J has length greater than 3/4. Using the same logic as in step 1, we can calculate that the probability that interval J has length greater than 3/4 is also  1/4.

Multiply the two probabilities to find the probability that both intervals I and J have length greater than 3/4. Since x1​, x2​, x3​, and x4​ are chosen independently at random, the events that interval I has length greater than 3/4 and interval J has length greater than 3/4 are independent. Therefore, the probability that both intervals I and J have length greater than 3/4 is the product of the two probabilities we calculated in steps 1 and 2, which is 1/4 * 1/4 = 1/16.

Oct 9, 2023
edited by Alan  Oct 10, 2023
#2
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.As Melody often notes, it can be useful to plot these sort of probabilities against each other.  Consider x_1 and x_2:

The probabilities of the pair can be thought of as uniformly randomly scattered points within a unit square.

The only regions of the square in which the lengths between the two values are greater than 3/4, are within the two red triangles indicated in the image above.

The probability that the lengths are greater than 3/4 is therefore given by p = Area of triangles/Area of square

The same probability holds for the lengths between x_3 and x_4, so, because the two probabilities are independent, the overall probability is given by p^2.

I'll leave you to do the number crunching.

Oct 11, 2023
edited by Alan  Oct 11, 2023
edited by Alan  Oct 11, 2023
edited by Alan  Oct 12, 2023
edited by Alan  Oct 12, 2023