a halloween candy jar has 8 snickers and 5 crunch bars. If three are chosen

Question

Then,

How many ways can it be at least one crunch bar?

Answer 1

In probability if we have 2 events that are mutually exclusive (they can't both happen) then their probabilities sum to 1.

 

So what is the probability that NO crunch bars are selected?  This is the probability we select 3 snickers bars in a row!

 

If 8 of the 13 bars are snickers, then 8/13 is the probability we select a snickers first.  Then there are 7 left of 12 bars, so 7/12 is the probability we get a second snickers, and 6/11 is the probability the third is a snickers.

 

The probability of these three events occurring in a row is the product of the probabilities, so prob of 3 snickers = 8*7*6/(13*12*11).

 

The probability of at least one crunch bar is 1-8*7*6/(13*12*11).