In a box of 12 pens, a total of 3 are defective. if a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective?

Answer in fraction form: 6/11
Answer in decimal form: 0.54545
(note: the answer in decimal form is approximate)



There are 12 pens total. Assuming we select without replacement, there are 12 ways to pick the first selection and then 11 to make the second selection. So there are 12*11 = 132 ways to pick two pens from a pool of twelve. This is if order mattered. Since order doesn't matter we divide by 2 to get 132/2 = 66

There are 66 ways to pick two pens out of a pool of twelve. This is where order does NOT matter.
We'll use this value later. Call it N = 66

We have 12 pens, of which 3 are defective. So 12-3 = 9 non-defective pens. 
There are 9 ways to pick the first non-defective pen. Then 8 ways to pick the second non-defective pen
Making 9*8 = 72 ways to pick the two non-defective pens if order mattered
Again order doesn't matter so 72/2 = 36 is the total number of ways to pick the two non-defective pens.
Call this value M = 36

Now divide M over N
M/N = 36/66 = (6*6)/(6*11) = 6/11

The probability in fraction form is 6/11
Using a calculator, we get 6/11 = 0.54545 which is approximate

This does not make sense

20000 as first it increases by 2500 then by 5000 so next it will increase by 10000