Question: A geotechnical engineer is analyzing soil samples from three different boreholes, each sample returned a measurement of slope stability ranging from 1 to 6 (inclusive), like a die roll. What is the probability that the sum of the three measurements is even? - NBX Soluciones
SEO Article: What Is the Probability That the Sum of Three Soil Slope Stability Measurements Is Even?
SEO Article: What Is the Probability That the Sum of Three Soil Slope Stability Measurements Is Even?
When a geotechnical engineer analyzes soil samples from three boreholes—each yielding a slope stability value from 1 to 6 (inclusive), similar to rolling a fair six-sided die—one engaging question naturally arises: What is the probability that the sum of the three measurements is even? Understanding this probability helps engineers assess consistency and risk in subsurface stability assessments.
Understanding the Problem
Understanding the Context
Each borehole produces a slope stability reading between 1 and 6, inclusive. Since the die is fair, each number from 1 to 6 has an equal probability of 1/6. The problem asks for the likelihood that the sum of three independent readings is an even number.
Mathematically, we want:
> P(sum of three numbers is even)
Because each sample is of a number from 1 to 6, we begin by analyzing the parity (odd or even nature) of each die roll.
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Key Insights
Step 1: Parity of Each Die Roll
The numbers 1 to 6 consist of:
- Even numbers: 2, 4, 6 → 3 outcomes
- Odd numbers: 1, 3, 5 → 3 outcomes
Each outcome has probability 3/6 = 1/2. So each reading independently has a 50% chance of being odd and 50% of being even.
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Step 2: Parity of the Sum
The sum of three numbers is even if:
- All three numbers are even, or
- Exactly two numbers are odd and one even (odd + odd + even = even + even = even)
Let’s compute the probabilities of these two cases.
Case 1: All three even
Probability = (1/2) × (1/2) × (1/2) = 1/8
Case 2: Two odd, one even
There are 3 ways this can happen (odd-odd-even, odd-even-odd, even-odd-odd)
Probability for one such combination (e.g., odd-odd-even) = (1/2) × (1/2) × (1/2) = 1/8
Total for all 3 combinations = 3 × (1/8) = 3/8
