Probability of Having at Least Two Left-Handed People in a Group of 12
Understanding how to calculate probabilities, especially in scenarios involving a specific number of individuals with certain traits, is a fundamental concept in statistics. This article will walk you through the process of calculating the probability that at least two people in a randomly chosen group of 12 are left-handed, given that 7.5% of the population is left-handed. We will use the binomial distribution to find the solution.
Introduction to Binomial Distribution
The binomial distribution is a probability distribution that models the number of successes in a fixed number of independent Bernoulli trials. Each trial has only two possible outcomes: success or failure. In this context, 'success' can be defined as selecting a left-handed person from a given total population.
Defining Parameters
Let's define the parameters for our scenario:
- n (the number of people in the group) 12.
- p (the probability of a randomly chosen person being left-handed) 0.075.
- q (the probability of a person not being left-handed) 1 - 0.075 0.925.
Calculating Probabilities
We need to find the probability that there are at least two left-handed people in the group. This can be calculated using the binomial distribution formula. The formula for the binomial probability is:
P(X k) C(n, k) * p^k * q^(n-k), where C(n, k) is the binomial coefficient.
Probability of Exactly 0 Left-Handed People
The probability of having exactly 0 left-handed people in the group is given by:
P(X 0) C(12, 0) * 0.075^0 * 0.925^12
P(X 0) 1 * 1 * 0.925^12
P(X 0) ≈ 0.925^12
P(X 0) ≈ 0.392
Probability of Exactly 1 Left-Handed Person
The probability of having exactly 1 left-handed person in the group is given by:
P(X 1) C(12, 1) * 0.075^1 * 0.925^11
P(X 1) 12 * 0.075 * 0.925^11
P(X 1) ≈ 0.384
Combining Probabilities
To find the probability of having at least 2 left-handed people, we need to subtract the probabilities of having exactly 0 and exactly 1 left-handed person from 1:
P(X ≥ 2) 1 - P(X 0) - P(X 1)
P(X ≥ 2) 1 - 0.392 - 0.384
P(X ≥ 2) ≈ 1 - 0.776
P(X ≥ 2) ≈ 0.224
Conclusion
The probability of finding at least 2 left-handed people in a randomly chosen group of 12 is approximately 0.224, or 22.4%. This calculation is crucial in understanding probabilities in diverse scenarios and is a foundational concept in statistics and probability theory.
Additional Insight
To further understand the concept, let's consider a similar scenario using 30 people. The probability that at least one of 30 people selected is left-handed can be calculated as follows:
P(one of 30 people selected is left-handed) 1 - 30C0 * 0.075^0 * 0.925^30
P(one of 30 people selected is left-handed) 1 - 0.925^30
P(one of 30 people selected is left-handed) ≈ 1 - 0.287 ≈ 0.9036
Similarly, for 2 left-handed people in 30:
P(two of 30 people selected are left-handed) 30C2 * 0.075^2 * 0.925^28
P(two of 30 people selected are left-handed) ≈ 21