Question

Fred needs to print his term paper. He pulls down a box of twelve ink cartridges and recalls that three of them have no more ink. If he selects four cartridges from the box, what is the probability that (a) no cartridge has ink (b) one cartridge has no ink (c) hree cartridges have no ink?

Answer 1

answer to A is evidently zero, because if he picks out four cartridges, at least one has ink (since only three are empty)

Answer 2

lol that is the only one i figured out.

Answer 3

one cartridge has no ink means 3 do and one does not. that would be
\[\frac{\dbinom{9}{3}\times\dbinom{3}{1}}{\dbinom{12}{3}}\]

Answer 4

which does't really answer the question, just tell you what you need to compute.
numerator means number of ways to choose 3 out of the 9 good cartridge, times the number of ways to choose one bad one
denominator is the total number or ways to choose 4 from a set of 12, so now i see i made a mistake

Answer 5

\[\frac{\dbinom{9}{3}\times\dbinom{3}{1}}{\dbinom{12}{4}}\] that's better

Answer 6

\[\dbinom{9}{3}=\frac{9\times 8\times 7}{3\times 2}=3\times 7\times 7=84\]
\[\dbinom{3}{1}=3\]
\[\dbinom{12}{4}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2}=11\times 5\times 9=495\]

Answer 7

thank you