For this problem, assume that three students are chosen from 5 freshmen, 4 sophomores, and 4 juniors. what is the probability of selecting 2 freshmen and 1 junior given that at least one freshman will be selected?

The formula , where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

The sample space S in our problem is as follows: 

triples selected from 5 freshmen, 4 sophomores, and 4 juniors, so that at least 1 freshman is selected.

The selection of 1 freshman can be done in 5 ways. The selection of the remaining 2 out of 4 freshmen (one is already chosen), 4 sophomores, and 4 juniors can be done in C(12, 2) many ways.

So there are in total, 5*66= 330 possible triples.

n(S)=330

The event E "selecting 2 freshmen and 1 junior" can occur in :

many ways.

so n(E)=40

P(E)=n(E)/n(S)=40/330=0.12

0.12

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,