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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 $$J$$ overlap.

May 1, 2024

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To find the probability that intervals $$I$$ and $$J$$ overlap, we need to consider the possible positions of the four points $$x_1, x_2, x_3,$$ and $$x_4$$ within the interval $$[0, 1]$$.

Without loss of generality, let's assume that $$x_1 < x_2$$ and $$x_3 < x_4$$.

For $$I$$ and $$J$$ to overlap, one of the following conditions must be true:

1.  $$x_1 < x_3 < x_2 < x_4$$

2.  $$x_3 < x_1 < x_4 < x_2$$

Let's calculate the probability for each condition:

1.  Probability of Condition 1:

- The probability that $$x_1$$ falls in the interval $$[0, 1]$$ is $$1$$.

- The probability that $$x_3$$ falls in the interval $$[x_1, 1]$$ is $$1 - x_1$$.

- Given $$x_1$$ and $$x_3$$, the probability that $$x_2$$ falls in the interval $$(x_1, 1]$$ is $$1 - x_1$$.

- Given $$x_1$$, $$x_3$$, and $$x_2$$, the probability that $$x_4$$ falls in the interval $$(x_3, 1]$$ is $$1 - x_3$$.

- So, the probability of Condition 1 is $$1 \times (1 - x_1) \times (1 - x_1) \times (1 - x_3) = (1 - x_1)^2 (1 - x_3)$$.

2.  Probability of Condition 2:

- The probability that $$x_3$$ falls in the interval $$[0, 1]$$ is $$1$$.

- The probability that $$x_1$$ falls in the interval $$[0, x_3]$$ is $$x_3$$.

- Given $$x_3$$ and $$x_1$$, the probability that $$x_4$$ falls in the interval $$(x_3, 1]$$ is $$1 - x_3$$.

- Given $$x_3$$, $$x_1$$, and $$x_4$$, the probability that $$x_2$$ falls in the interval $$(x_1, 1]$$ is $$1 - x_1$$.

- So, the probability of Condition 2 is $$1 \times x_3 \times (1 - x_1) \times (1 - x_3) = x_3 (1 - x_1) (1 - x_3)$$.

Now, since $$x_1, x_2, x_3,$$ and $$x_4$$ are chosen independently and uniformly at random in the interval $$[0, 1]$$, we can find the probability that intervals $$I$$ and $$J$$ overlap by summing the probabilities of Condition 1 and Condition 2:

$\text{Total probability} = (1 - x_1)^2 (1 - x_3) + x_3 (1 - x_1) (1 - x_3)$

$= (1 - x_1) (1 - x_3) (1 - x_1 + x_3)$

$= (1 - x_1) (1 - x_3)$

Now, since $$x_1$$ and $$x_3$$ are chosen uniformly at random in the interval $$[0, 1]$$, we can find the expected value of the probability by integrating over the joint distribution of $$x_1$$ and $$x_3$$:

$\text{Expected probability} = \int_{0}^{1} \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_1 \, dx_3$

$= \int_{0}^{1} \left( \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_3 \right) dx_1$

$= \int_{0}^{1} \left( 1 - x_1 - \frac{1}{2} + \frac{x_1^2}{2} \right) dx_1$

$= \left[ x_1 - \frac{x_1^2}{2} - x_1 + \frac{x_1^3}{6} \right]_{0}^{1}$

$= \left( 1 - \frac{1}{2} - 1 + \frac{1}{6} \right) - \left( 0 - 0 - 0 + 0 \right)$

$= \frac{1}{3} - \frac{1}{2} = \frac{1}{6}$

Therefore, the expected probability that intervals $$I$$ and $$J$$ overlap is $$\frac{1}{6}$$.

May 1, 2024