Probabilities
Conditions:
In a class of 28 students, 12 are dual-enrolled (DE) and 16 are not dual-enrolled (NDE).
We randomly select two students from the class without replacement.
We'll use a tree diagram. Let's write it down below:
We can select one of the combinations: {DE/DE, DE/NDE, NDE/DE, NDE/NDE}
In the first selection, there are 12 NDE and 16 DE, thus the probabilities are:
P(DE) = 16/28 and P(NDE) = 12/28
If we select one DE, there will be only 15 DE's and 12 NDE, thus the second selection will have the probabilities:
P(DE) = 15/27 and P(NDE) = 12/27
Now assume the first selection was NDE, thus the second selection has the following probabilities:
P(DE) = 16/27 and P(NDE) = 11/27
The probability of both students being dual-enrolled is:
P(DE,DE) = 16/28*15/27 = 20/63
The probability of the first student being dual-enrolled and the second student not dual-enrolled is:
P(DE,NDE) = 16/28 * 12/27 = 16/63
Riounding the answer to three decimal places:
P(DE,DE) = 0.317
P(DE,NDE) = 0.254