Please Help, Geometric Probability

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 $\frac{1}{6}$.