Defective battery problem 
A flashlight has $6$ batteries, $3$ of which are defective. If $3$ are selected at random without replacement, find the probability that all of them are defective.

I am finding the probability of getting all of them defective batteries which should be the probability of each when its drawn, should this be like this: $(3/6) \cdot (2/5) \cdot (1/4) = 6/120 = 0.05$. When submitting this i get an error, but isn't finding the probability like this is basically finding the probability for each with order is finding it for all of them?
 A: Your calculation is sound. You can get the same answer from considering the binomial coefficient $\binom 63 = \frac{6!}{3!(6-3)!} = 20$ (meaning here there are $20$ different ways to chose $3$ items from $6$) and observing that there is only one of these options that results in choosing all the defective batteries, giving a probability of $\frac {1}{20}$ as you found.
A: If three out of the six are defective and you select three without replacement, there is only one way to obtain all three defective batteries.  But there are clearly more ways to select three batteries in which one or more is not defective.
To see this, it suffices to label the batteries as follows:
$$\{G_1, G_2, G_3, D_1, D_2, D_3\}.$$  Then there are $$\binom{6}{3} = \frac{6!}{3!3!} = 20$$ ways to select three batteries without replacement.  But only one way gives you $\{D_1, D_2, D_3\}$.  The full list of $20$ possibilities is:
$$\{G_1, G_2, G_3\}, \{G_1, G_2, D_1\}, \{G_1, G_2, D_2\}, \{G_1, G_2, D_3\}, \{G_1, G_3, D_1\}, \\
\{G_1, G_3, D_2\}, \{G_1, G_3, D_3\}, \{G_1, D_1, D_2\}, \{G_1, D_1, D_3\}, \{G_1, D_2, D_3\}, \\
\{G_2, G_3, D_1\}, \{G_2, G_3, D_2\}, \{G_2, G_3, D_3\}, \{G_2, D_1, D_2\}, \{G_2, D_1, D_3\}, \\
\{G_2, D_2, D_3\}, \{G_3, D_1, D_2\}, \{G_3, D_1, D_3\}, \{G_3, D_2, D_3\}, \{D_1, D_2, D_3\}$$
