What is the probability that she chose the four students so that exactly one of them is a junior?

  B)  0.246

Step-by-step explanation:

The desired probability is the sum of the probabilities of each of the ways one junior can be chosen. If we let a, b, c, d represent the ratio of juniors in classes 1 to 4, then this probability is ...

  p(1 junior) = (a)(1-b)(1-c)(1-d) + (1-a)(b)(1-c)(1-d) + (1-a)(1-b)(c)(1-d) + (1-a)(1-b)(1-c)(1-d)

We can simplify this a little bit to ...

  p(1 junior) = (1-a)(1-b)(1-c)(1-d) × (a/(1-a) +b/(1-b) +c/(1-c) +d/(1-d))

___

So, for {a, b, c, d} = {9/15, 12/15, 6/15, 3/15} = {3/5, 4/5, 2/5, 1/5} the probability of interest is ...

  p(1 junior) = (2/5·1/5·3/5·4/5) × (3/2 + 4/1 + 2/3 + 1/4) = 24/625 × 77/12

  p(1 junior) = 154/625 ≈ 0.246

Comment on the sum we're computing

Each of the probabilities in the first sum above is ...

  p(junior in given class) · p(not a junior in the other 3 classes)

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?