A batch of 10 gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets.

Question

kropot72 · Answer

$$P(2\ defective)=\frac{4C2\times6}{10C3}=\frac{6\times6\times6}{10\times9\times8}=you\ can\ calculate$$

anonymous · Answer

Where 6 came from? and what is C?

kropot72 · Answer

'C' is an abbreviation that stands for Combinations or for Choose. For example:
4C2 means the number of combinations of four things taken two at a time, or alternatively the number of ways of choosing two things out of four things.
Have you studied combinations?
There are six good gaskets. The number of combinations of the six good gaskets taken one at a time is represented by 6C1 which equals 6.