Rework problem 5 from section 2.4 of your text (page 81) about randomly selecting 3 cans of regular and diet cola, but assume that there are 6 cans of regular cola and 8 cans of diet cola. what is the probability that 2 cans of regular cola and 1 can of diet cola are 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.

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There are 6 cans of regular cola, and 8 cans of diet cola.

in short, let's use the letters R, for regular, and D for diet cola.

So we have 6 R and 8 D. 14 in total.

There are C(14, 3) many ways of picking 3 cans out of 14, so there are 



ways of forming groups of 3 cans.

Consider the event "2 cans of regular cola and 1 can of diet cola are selected", 

this may happen in C(6, 2)*C(8, 1) many ways, since there are C(6, 2) ways of selecting 2 regular colas out of 6, which can be combined with C(8, 1) ways of selecting 1 diet cola out of 8.



P(2 cans of regular cola and 1 can of diet cola)=120/364=0.33

0.33