Probability of selecting all defective parts In a problem we are faced with 22 items from a shipment. It is known that 7 of the items are defective. What is the probability of selecting all defective items from a draw of 7. What is the probability of a draw where none of the items defective. I suspect both answers are the same but I could very well be wrong.
The way I see it -
P(E)=(P(one Defective)=7/22=0.3181818182
P(E)=(P(all defective)^7)=0.0003301623
Am I even approaching this problem from the right direction?
 A: There are $\binom{22}{7} = 170,544$ ways to draw seven parts.  There is only one way to get all parts being defective (you have to draw every last one as being defective: $\binom{7}{7} = 1$).
On the other hand, there are $22 - 7 = 15$ non-defective parts, so there are $\binom{15}{7} = 6,435$ ways to draw all non-defective parts.
Additional Details
This can be done by either permutations or combinations although combinations is generally more compact and intuitive and thus you should try to think in terms of combinations for most problems--but not all problems.
To do this using permutations, we note that you have a $\frac{7}{22}$ chance of drawing a defective part, then it's $\frac{6}{21}$, then $\frac{5}{20}$, etc.  This gives:
$$
p = \frac{7}{22}\cdot\frac{6}{21}...\cdot\frac{1}{16} = \frac{7!\cdot15!}{22!}
$$
However, you will notice that this equals the explicit form of $\frac{1}{\binom{22}{7}}$:
$$
\binom{n}{k} = \frac{n!}{k!(n - k)!}
$$
We then would get:
$$
\frac{1}{\binom{22}{7}} = \frac{1}{\frac{22!}{7!(22 - 7)!}} = \frac{7!\cdot 15!}{22!}
$$
For getting all seven as being non-defective then we get:
$$
p = \frac{15}{22}\cdot\frac{14}{21}...\frac{9}{16} = \frac{15!\cdot15!}{22!\cdot 8!}
$$
Again, if we write the above combinations, we'll get the same thing:
$$
\frac{\binom{15}{7}}{\binom{22}{7}} = \frac{7!\cdot15!}{22!}\cdot \frac{15!}{7!(15 - 7)!} = \frac{15!\cdot15!}{22!\cdot8!}
$$
