Probability Of Subinterval Length Exceeding 0.4 In A Unit Interval

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This article delves into a fascinating probability problem concerning the partitioning of a unit interval. Imagine you have a line segment of length 1. Now, you randomly select three points on this segment, effectively dividing it into four subintervals. The core question we aim to answer is: What is the probability that at least one of these subintervals has a length greater than 0.4? This problem elegantly combines concepts of probability, inequalities, and geometric constraints, making it a captivating exploration for anyone interested in mathematical problem-solving.

To restate the problem clearly, consider a unit interval, which we can represent as the range [0, 1] on the number line. We independently and uniformly choose three points, let's call them x, y, and z, within this interval. This means each point has an equal chance of landing anywhere between 0 and 1. These three points will naturally partition the unit interval into four subintervals. Our objective is to determine the probability that at least one of these subintervals has a length exceeding 0.4.

To approach this problem systematically, we first need to establish a clear notation and framework. Let's assume, without loss of generality, that the three chosen points are ordered such that 0 ≤ x ≤ y ≤ z ≤ 1. This ordering simplifies our analysis by defining the subintervals in a consistent manner. The lengths of the four subintervals can then be expressed as follows:

  1. Length of the first subinterval: x - 0 = x
  2. Length of the second subinterval: y - x
  3. Length of the third subinterval: z - y
  4. Length of the fourth subinterval: 1 - z

Our main task now is to calculate the probability that at least one of these four lengths is greater than 0.4. This seemingly simple question requires a careful consideration of the possible values of x, y, and z, and how they interact to determine the lengths of the subintervals. Before we dive into the detailed solution, it's worthwhile to think about some initial approaches and potential challenges. One key strategy is to consider the complementary probability – the probability that none of the subintervals are longer than 0.4 – and subtract it from 1. This approach can often simplify calculations in probability problems.

To solve this problem effectively, we'll employ a geometric probability approach. This involves visualizing the possible values of x, y, and z as points within a three-dimensional space and then determining the region that corresponds to the event we're interested in. Since we've established the order 0 ≤ x ≤ y ≤ z ≤ 1, we can represent the possible values of (x, y, z) as points within a unit cube in the first octant, specifically the region defined by these inequalities. This region is a tetrahedron with vertices at (0, 0, 0), (0, 0, 1), (0, 1, 1), and (1, 1, 1). The volume of this tetrahedron represents the total possible outcomes, and it's equal to 1/6.

Now, we need to identify the region within this tetrahedron that corresponds to the event where none of the subintervals have a length greater than 0.4. This means we need to satisfy the following four inequalities simultaneously:

  1. x ≤ 0.4
  2. y - x ≤ 0.4
  3. z - y ≤ 0.4
  4. 1 - z ≤ 0.4, which implies z ≥ 0.6

These inequalities define a smaller region within the tetrahedron, and our goal is to calculate the volume of this smaller region. This volume will represent the probability that none of the subintervals exceed 0.4. To find this volume, we need to carefully analyze the geometry of the region defined by these inequalities. The region is also a tetrahedron, but its vertices are determined by the intersections of the planes defined by the inequalities. Finding these intersection points and setting up the appropriate integrals to calculate the volume can be a bit intricate, but it's a crucial step in solving the problem.

Once we've calculated the volume of this smaller tetrahedron, we can subtract it from the volume of the larger tetrahedron (1/6) to find the probability that at least one subinterval exceeds 0.4. This is because the probability of an event happening is 1 minus the probability of the event not happening. By focusing on the complementary event, we often simplify the calculations.

As discussed, we will calculate the complementary probability, which is the probability that none of the subintervals are longer than 0.4. This means the following conditions must all be true:

  1. x ≤ 0.4: The first subinterval's length (x) must be less than or equal to 0.4.
  2. y - x ≤ 0.4: The second subinterval's length must be less than or equal to 0.4.
  3. z - y ≤ 0.4: The third subinterval's length must be less than or equal to 0.4.
  4. 1 - z ≤ 0.4: The fourth subinterval's length must be less than or equal to 0.4, which implies z ≥ 0.6.

These four inequalities define a region in the three-dimensional space of (x, y, z). To visualize this region, we can think of each inequality as defining a half-space, and the intersection of these half-spaces is the region we're interested in. This region is a tetrahedron, and we need to find its vertices to calculate its volume. The vertices are the points where three or more of the boundary planes intersect.

Let's find these vertices:

  • Vertex 1: Intersection of x = 0.4, y - x = 0.4, and z - y = 0.4. Solving this system of equations gives us (x, y, z) = (0.4, 0.8, 1.2). However, since z cannot exceed 1, this vertex is outside our region of interest. We need to consider the intersection with z = 1 instead. So, intersecting x = 0.4, y - x = 0.4, and z = 1 gives us (x, y, z) = (0.4, 0.8, 1). This is one valid vertex.
  • Vertex 2: Intersection of x = 0.4, y - x = 0.4, and z = 0.6. Solving this gives (x, y, z) = (0.4, 0.8, 0.6). This point does not satisfy the condition z - y ≤ 0.4 because 0.6 - 0.8 = -0.2, which is ≤ 0.4, but we will consider the intersection of x = 0.4, z - y = 0.4, and z = 0.6 instead. The solution to this system is (x, y, z) = (0.4, 0.2, 0.6). This is another valid vertex.
  • Vertex 3: Intersection of x = 0.4, y - x = 0.4, and 1 - z = 0.4 (or z = 0.6). This was handled above.
  • Vertex 4: Intersection of x = 0.4, z - y = 0.4, and 1 - z = 0.4 (or z = 0.6). We already know x = 0.4 and z = 0.6 so y = 0.2. This gives us the point (0.4, 0.2, 0.6). This is vertex 2.
  • Vertex 5: Intersection of y - x = 0.4, z - y = 0.4, and 1 - z = 0.4 (or z = 0.6). Solving gives z = 0.6, y = 0.2 and x = -0.2. Which is outside the bounds.

After carefully considering all the intersections, we find the vertices of the tetrahedron to be:

(0.4, 0.8, 1) (0.4, 0.2, 0.6) (0, 0, 0.6) (0, 0.4, 1)

The vertices of the smaller tetrahedron representing the region where none of the subintervals exceed 0.4 are: (0.4, 0.8, 1), (0, 0.4, 1), (0, 0, 0.6) and (0.4,0.2,0.6).

The volume of a tetrahedron with vertices a, b, c, and d is given by:

Volume = (1/6) |(a - d) · ((b - d) × (c - d))|

Let us substitute these vertices into the formula where

a = (0.4, 0.8, 1) b = (0, 0.4, 1) c = (0, 0, 0.6) d = (0.4, 0.2, 0.6)

Then

a - d = (0, 0.6, 0.4) b - d = (-0.4, 0.2, 0.4) c - d = (-0.4, -0.2, 0)

(b - d) × (c - d) = | i j k -0.4 0.2 0.4 -0.4 -0.2 0 |

= i (0 - (-0.08)) - j (0 - (-0.16)) + k (0.08 - (-0.08)) = (0.08)i - (0.16)j + (0.16)k = (0.08, -0.16, 0.16)

(a - d) · ((b - d) × (c - d)) = (0, 0.6, 0.4) · (0.08, -0.16, 0.16) = (0) + (-0.096) + (0.064) = -0.032

Therefore Volume = (1/6)|-0.032| = 0.032/6 = 0.016/3

The volume of the region representing all possible outcomes, where 0 ≤ x ≤ y ≤ z ≤ 1, is 1/6. Therefore, the probability that none of the subintervals exceed 0.4 is:

P(none > 0.4) = (Volume of tetrahedron) / (Volume of the entire region) = (0.016/3) / (1/6) = 0.032.

Finally, to answer our original question, we subtract the complementary probability from 1 to find the probability that at least one subinterval is longer than 0.4:

P(at least one > 0.4) = 1 - P(none > 0.4) = 1 - 0.032 = 0.968.

Therefore, the probability that at least one of the four subintervals formed by three independently and uniformly chosen points on a unit interval is longer than 0.4 is 0.968. This problem demonstrates the power of geometric probability in solving seemingly complex problems by translating them into geometric representations. By carefully considering the constraints and using geometric arguments, we can arrive at a precise and insightful solution.

Probability, Subinterval Length, Unit Interval Partition, Geometric Probability, Inequalities, Constraints, Complementary Probability, Tetrahedron Volume, Uniform Distribution, Independent Events. These keywords are crucial for understanding the core concepts involved in the problem and for searching for related topics and information.