Out of 50 students, X=30, Y=15, making up 90% of the students.
Tree diagram consists at the top of the 50 students, branching out to the second level 30, 15 and 5 for the three disciplines. The project team tree has three branches but, because students are randomly chosen, we cannot say what disciplines they represent, but we can represent the probabilities as another level of branches 1/10, 3/5 and 3/10 corresponding to 10%, 60% and 30% attached to each of the branches for the first level branches. (Strictly speaking, the selection process may change the probabilities, because the first student selected increases the probabilities for the remaining students by reduction of the pool of students to 49. If, for example, an X student was the first to be selected for the project team, then out of the 49 remaining, there would be 29 X students, changing the probabilities to 29/49, 15/49 and 5/49.)
The joint probability distribution can be represented binomially as (p+(1-p))^3=p^3+3p^2(1-p)+3p(1-p)^2+(1-p)^3, where p=0.90 (for combined decision sciences and industrial statistics majors). So 1=0.729+0.243+0.027+0.001. The terms in this series separately show the make up of the project team as probabilities of all X or Y, 2 X or Y plus a business mathematics student, 2 of the latter and one X or Y, or all business maths students,
If X=1 then 49 students remain, consisting of 29 X type, 15 Y type and 5 business maths majors. The binomial distribution of selecting Y is given by (p1-(1-p1))^2=p1^2+2p1(1-p1)+(1-p1)^2, where p1=15/49, the probability of selecting a Y student, and 1-p1=34/49 is the probability of selecting another X or a business maths student. So 1=0.0937+0.4248+0.4815. That means 9.37% probability of 2 Y students, 42.48% probability of 1Y and either of the other two types, and 48.15% probability of no Y type students.
The mean in a binomial distribution is np, where n in this case is 3 (size of the project team) and p=3/5 or 0.6, so the mean is 1.8. This assumes that the mean applies before any member of the team has been selected, rather than when X=1, after a selection has already been made.
For Y the mean is 3*15/49=0.92 approx., given that X=1.
Before any member of the team has been selected, the probabilities are independent. The only dependence X and Y share is that once a team member has been selected, and is found to be X, for example, the probability changes because there is one fewer of that type left in the pool of students. Read More: ...