What is the probability of the class having 2 left-handed students?
To determine the probability of the class having 2 left-handed students, we need to know the total number of students in the class and the proportion of left-handed students in the population.
Let's assume there are 30 students in the class, and the proportion of left-handed students in the population is 10%.
To calculate the probability, we can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting k successes (in this case, having 2 left-handed students)
- C(n, k) is the number of ways to choose k successes out of n trials (combination)
- p is the probability of success in one trial (proportion of left-handed students)
- (1-p) is the probability of failure in one trial (proportion of right-handed students)
- n is the total number of trials (total number of students)
Using the given values, we have:
P(X = 2) = C(30, 2) * 0.10^2 * (1-0.10)^(30-2)
Calculating the values:
C(30, 2) = (30!)/(2!(30-2)!) = (30 * 29)/2 = 435
0.10^2 = 0.01
(1-0.10)^(30-2) = 0.9^28 ≈ 0.0912
Plugging the values back into the formula:
P(X = 2) ≈ 435 * 0.01 * 0.0912 ≈ 0.3972
Therefore, the probability of the class having 2 left-handed students is approximately 0.3972, or 39.72%.