Stats problem/ Hypergeometric distribution can't figure out the combinations

[mfdoom11]

Suppose there are 14 items in a lot. 2 are defective and 12 are non-defective. 6 are inspected. Find the following, rounded to 4 decimals.

a) Find the probability that all of those chosen are non-defective.

b) Find the probability that all of those chosen are defective.

c) Find the probability that at least 1 is defective.

Can you please explain what you plugged in and the answers

 

[Romsek]

You can't figure out any of it?

(a) is just \(\displaystyle \dfrac{\dbinom{12}{6}\dbinom{2}{0}}{\dbinom{12}{6}}\)

We choose 6 of the !defective items, of which there are 12, and 0 of the defective items, of which there are 2.
Then we normalize this by the total number of sets of 6 we could choose from all 12 parts.

(b) is rather impossible unless you're saying items are inspected and replaced before selecting the next. This is a very different problem if so.

(c) the probability at least 1 is defective is 1 minus the probability 0 are defective, see part (a)

 

[pka]

Suppose there are 14 items in a lot. 2 are defective and 12 are non-defective. 6 are inspected. Find the following, rounded to 4 decimals.
b) Find the probability that all of those chosen are defective.

In part b) Do you mean that both defective parts are among the six inspected?