Final answer:
The probability that at least 3 out of 5 students will get version A of the test is approximately 34.0%.
Step-by-step explanation:
To find the probability that at least 3 out of 5 students will get version A of the test, we can use the binomial probability formula. Let's consider the opposite scenario where less than 3 students get version A. The probability of a student getting version A is 1/4. So the probability of a student not getting version A is 1 - 1/4 = 3/4.
To find the probability of less than 3 students getting version A, we can calculate:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
where X follows a binomial distribution with n = 5 (number of students) and p = 1/4 (probability of getting version A).
Using the binomial probability formula, the probability of getting exactly x successes in n trials is:
P(X = x) = (nCx)(p^x)((1 - p)^(n - x))
where nCx is the number of combinations of n items taken x at a time.
Calculating the probabilities for X = 0, 1, and 2, we get:
P(X = 0) = (5C0)((1/4)^0)((3/4)^(5 - 0)) = 1(1)(1) = 1
P(X = 1) = (5C1)((1/4)^1)((3/4)^(5 - 1)) = 5(1/4)(81/256) = 405/1024
P(X = 2) = (5C2)((1/4)^2)((3/4)^(5 - 2)) = 10(1/16)(27/64) = 270/1024
Adding up these probabilities:
P(X < 3) = 1 + 405/1024 + 270/1024 = 675/1024
Finally, to find the probability that at least 3 out of 5 students will get version A:
P(X >= 3) = 1 - P(X < 3) = 1 - 675/1024 = 349/1024 ≈ 34.0%