Math Problem Statement

What is the probability that two numbers selected at random between 0 and 10 have a sum greater than or equal to 8 if the numbers are not necessarily integers?

Solution

We are tasked with finding the probability that two random numbers selected between 0 and 10 have a sum greater than or equal to 8. The numbers are not restricted to being integers, meaning they are continuous random variables.

Step 1: Define the sample space

The sample space consists of all possible pairs of numbers (x,y)(x, y) where both xx and yy lie between 0 and 10. Geometrically, this corresponds to all points in a 10x10 square on the coordinate plane, where: 0x10and0y10.0 \leq x \leq 10 \quad \text{and} \quad 0 \leq y \leq 10. The total area of this square is: Area of the sample space=10×10=100.\text{Area of the sample space} = 10 \times 10 = 100.

Step 2: Define the favorable region

We are interested in the region where the sum of the two numbers is greater than or equal to 8, i.e., x+y8x + y \geq 8. This inequality represents a line on the coordinate plane: x+y=8.x + y = 8. The region where x+y8x + y \geq 8 lies above and to the right of this line, bounded by x=0x = 0, y=0y = 0, x=10x = 10, and y=10y = 10.

The line x+y=8x + y = 8 intersects the xx-axis at x=8x = 8 and the yy-axis at y=8y = 8. The region of interest forms a right triangle with vertices at (8,0)(8, 0), (0,8)(0, 8), and (10,0)(10, 0), plus the area to the top-right of the triangle.

Step 3: Compute the favorable area

  1. Area of the triangle: The right triangle formed by the line x+y=8x + y = 8 with the xx-axis and yy-axis has base and height of length 8. The area of the triangle is: Area of the triangle=12×8×8=32.\text{Area of the triangle} = \frac{1}{2} \times 8 \times 8 = 32.
  2. Remaining rectangular area: There is a rectangular region above the triangle with a width of 108=210 - 8 = 2 (between x=8x = 8 and x=10x = 10) and height 10. The area of this rectangle is: Area of the rectangle=2×10=20.\text{Area of the rectangle} = 2 \times 10 = 20.

So, the total area of the region where x+y8x + y \geq 8 is: Favorable area=32+20=52.\text{Favorable area} = 32 + 20 = 52.

Step 4: Calculate the probability

The probability is the ratio of the favorable area to the total area of the sample space: P(x+y8)=52100=0.52.P(x + y \geq 8) = \frac{52}{100} = 0.52.

Thus, the probability that two randomly selected numbers between 0 and 10 have a sum greater than or equal to 8 is 0.520.52 or 52%.

Would you like more details or have any questions?

Related Questions:

  1. What is the probability that the sum of the two numbers is less than 5?
  2. How would the result change if the numbers were integers instead of continuous?
  3. What is the probability that two random numbers between 0 and 1 have a sum greater than or equal to 0.5?
  4. What is the expected value of the sum of two random numbers between 0 and 10?
  5. How do we compute probabilities for sums of random variables over other intervals?

Tip:

When dealing with continuous random variables, probabilities often correspond to areas under curves or in geometric regions.

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Math Problem Analysis

Mathematical Concepts

Probability
Geometry
Continuous Random Variables

Formulas

Area of a triangle: A = 1/2 * base * height
Probability formula: P = favorable area / total area

Theorems

Geometric Probability Theorem

Suitable Grade Level

Grades 10-12