Fred needs to print his term paper. He pulls down a box of 12 ink cartridges and recalls that 3 of them have no more ink. If he selects 4 cartridges from the box, find the probability that a) one cartridge has no ink and b) 3 cartridges have no ink.

Question

Fred needs to print his term paper.  He pulls down a box of 12 ink cartridges and recalls that 3 of them have no more ink.  If he selects 4 cartridges from the box, find the probability that a) one cartridge has no ink and b) 3 cartridges have no ink.

anonymous · Answer

Ok, so this is just like the one you did before. Can you give an idea how you should set it up?

anonymous · Answer

Well I tried by doing this.  4C1 = 4 for the four he selects and the 1 with no ink.  But that's as far as I got.

anonymous · Answer

For part a)

How many different ways can he choose 1 cartridge from the 3 that have no ink?

anonymous · Answer

3C1?

anonymous · Answer

Correct. Any how many different possible ways can he grab 4 cartridges from all the ones in the box?

anonymous · Answer

12C4.

anonymous · Answer

Correct.  So you divide your number of different successful outcomes by the number of total possible outcomes. The result is the probability.

anonymous · Answer

If I divide 3C1 (which equals 3) by 12C4 (which equals 495) I get 0.006 but that's not correct so I must be missing something.

jdcap · Answer

I see your post was two years ago, but I would like to tell you I am doing that exercise and you were right, you were missing something.
The probability is given by:
$$P(1-no ink cartridge)=\frac{ 3C3*9C1 }{ 12C4 } = \frac{ 28 }{ 55 }\approx0.51$$