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 \( \frac{1}{6} \).
