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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?

user140705
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    Hint : There are $\binom{22}{7}$ draws of $7$, but only one such that you select all defective items. Also, if you selected no defective item, then you selected items from the remaining $22-7 = 15$ others. – Amateur Apr 06 '14 at 01:10

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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:

$$ \frac{\binom{15}{7}}{\binom{22}{7}} = \frac{7!\cdot15!}{22!}\cdot \frac{15!}{7!(15 - 7)!} = \frac{15!\cdot15!}{22!\cdot8!} $$

Jared
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    @user140705 Note that your reasoning is incorrect. However, it can be made correct as follows : picking all $7$ defective items is equivalent to picking one at each one of the $7$ picks. But at the second pick, there are $21$ items left and $6$ defective items remaining if you picked one at first. Continuing this reasoning, one gets that the probability of picking the $7$ defective items is $\frac{7}{22} \times \frac{6}{21} \times \frac{5}{20} \times \frac{4}{19} \times \frac{3}{18} \times \frac{2}{17} \times \frac{1}{16}$. You can check it also equals $\frac{1}{\binom{22}{7}}$. – Amateur Apr 06 '14 at 01:19
  • Quick question. The answer above is confusing for ways to draw 7 non-defective items. Both calculations come up with 170544 ways. However, we know for a fact at least 1 way has all seven defective. Surely there are other possibilities for drawing 2 defective, or 3 defective, et cetera reducing the overall possible ways from 170544. – user140705 Apr 06 '14 at 01:25
  • @user140705 See my edited answer (I initially had $\binom{22}{15}$ which is not correct and makes no sense if you think about it). – Jared Apr 06 '14 at 01:26
  • No worries. I thought I had overlooked something. Thank you both greatly. – user140705 Apr 06 '14 at 01:27
  • @user140705 well you caught the obvious mistake: how could getting all non-defective have probability $1$? It doesn't pass the "smell" test (i.e. it's obviously incorrect which you noticed). – Jared Apr 06 '14 at 01:35