What is the probability of this? A shipment of 25 inexpensive digital watches, including 8 that are defective, is sent to a store. The store selects 6 at random for testing and rejects the whole shipment if 2 or more in the sample are found defective. What is the probability that the shipment will be rejected?
The answer in the back of the book is 0.6506
I just dont know how to get to it
 A: There are $\binom{25}{6}$ ways of choosing the sample.  You will reject unless  $0$ or $1$ defects are found.  The probability of accepting is therefore
$$
\frac{\binom{17}{6} \binom{8}{0}+ \binom{17}{5} \binom{8}{1}}{\binom{25}{6}} = \frac{442}{1265}
$$ 
and the probability of rejecting is $$\frac{823}{1265} \approx 0.650593$$
A: 
i've tried 8C2 * 17C4/ 25C6

This is the right idea, but that counts ways to obtain exactly 2 defectives.  
A shipment will be rejected if it contains at least 2 defectives.
Even better, the shipment is accepted if it has 0 or 1 defective; which is easier to count. The probability of being rejected is then obtained by the law of complementary probability.
$$1-\frac{{^8\mathrm C_0}{^{17}\mathrm C_6}+{^8\mathrm C_1}{^{17}\mathrm C_5}}{^{25}\mathrm C_{6}} = 1-\dfrac{61,880}{177,100}\approxeq 0.65059{\small 3}$$
A: If you have a sample of 6 and it is rejected then it has 2 or more defective watches. That is, it is composed of:
4 good ones, 2 bad
3 good, 3 bad
2 good, 4 bad
1 good, 5 bad
No good ones, 6 bad
Try to calculate in how many ways each of these can happen. 
Because we are mathematicians and because mathematicians are lazy we would rather calculate the likelihood of a sample being accepted and then calculate its complement.
The sample is accepted if there are no bad watches or only 1 watch.
How can you have a sample with no bad watches? Just take good watches from the shipment:
P(sample with no bad watches) = $\frac{17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20} $
Now what is the probability of the sample having only one bad watch? It is the likelihood of taking 5 good ones, one bad one, and then taking into account how many permutations of that action you can make. You can take a bad watch and 5 ones in 6 ways. It is one of:
$GGGGGB, GGGGBG, GGGBGG, GGBGGG, GBGGGG, BGGGGG $
So we get
P(sample with only one bad watch) = $6 \cdot\frac{8 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20}$
Calculating these, adding them up and then taking this out of 1, you get the desired result.
