Understanding the Probability of Sum of Three Random Numbers

In this article, we will delve into the concept of probability and how it applies to the sum of three random numbers generated within a specific interval. We will explore a detailed step-by-step process to determine the probability that the sum of three uniformly distributed random numbers in the interval [0, 2] falls between 1 and 2.

Defining the Problem

To solve this, we first need to define the problem clearly. We are seeking the probability that the sum of three random numbers, denoted as (X_1), (X_2), and (X_3), falls within the range [1, 2]. These numbers are independently and uniformly distributed over the interval [0, 2].

Total Volume of Possible Outcomes

The total space of possible outcomes for the three random numbers (X_1), (X_2), and (X_3) is the volume of the cube defined by the interval [0, 2]. This volume can be calculated as:

V_{text{total}} 2^3 8

Volume of the Favorable Region

To find the volume of the favorable region where (1 leq X_1 X_2 X_3 leq 2), we need to identify the intersection of the planes (X_1 X_2 X_3 1) and (X_1 X_2 X_3 2) with the cube [0, 2]^3.

Plane (X_1 X_2 X_3 1)

The vertices of the triangle formed in the (X_1 X_2 X_3) coordinate system are (1, 0, 0), (0, 1, 0), and (0, 0, 1). The area of this triangle is (frac{1}{2}).

Plane (X_1 X_2 X_3 2)

The vertices of the triangle formed are (2, 0, 0), (0, 2, 0), and (0, 0, 2). The area of this triangle is also (frac{1}{2}).

Volume Between the Planes

The volume of the region between these two planes can be calculated as the difference in volumes of the tetrahedrons formed. The volume of the tetrahedron for (X_1 X_2 X_3 leq 2) is:

V_2 frac{1}{6} cdot text{base area} cdot text{height} frac{1}{6} cdot 2 cdot 2 frac{4}{6} frac{2}{3}

The volume of the tetrahedron for (X_1 X_2 X_3 leq 1) is:

V_1 frac{1}{6} cdot text{base area} cdot text{height} frac{1}{6} cdot 1 cdot 1 frac{1}{6}

Therefore, the volume of the favorable region is:

V_{text{favorable}} V_2 - V_1 frac{2}{3} - frac{1}{6} frac{4}{6} - frac{1}{6} frac{3}{6} frac{1}{2}

Calculating the Probability

The probability that the sum of three random numbers generated in the interval [0, 2] is at least 1 and no more than 2 is given by the ratio of the favorable volume to the total volume of the cube:

P(1 leq X_1 X_2 X_3 leq 2) frac{V_{text{favorable}}}{V_{text{total}}} frac{frac{1}{2}}{8} frac{1}{16}

Therefore, the probability is:

Conclusion

The probability that the sum of three random numbers generated in the interval [0, 2] is at least 1 and no more than 2 is (boxed{frac{1}{16}}).

Related Keywords

keywords: probability, random numbers, uniform distribution