TITLE: Difficult probability of choosing ball from bag with $7$ balls labelled from $1-7$
QUESTION [4 upvotes]: This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with probability, which perhaps yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

A bag contains seven balls numbered from $1$ to $7$. A ball is chosen at random
  and its number is noted. The ball is then returned to the bag. This is done a total of seven times.
$(a)$ What is the probability that each ball is selected exactly once?
$(b)$ What is the probability that at least one ball is not selected?
$(c)$ What is the probability that exactly one of the balls is not selected? 

My thoughts:
$(a)\, \left(\frac 17\right)^7=\frac{1}{823543}$
$(b) \, 1 - \frac{1}{823543}?$
$(c)$ No idea on this one. My head aches after just thinking about it.

REPLY [2 votes]: Ad (a): You have calculated the probability that a specific order is selected. But there are $7!$ possible orders.
Ad (b): You are right to use the converse probability. Just use the hint for (a).
Ad (c): The result of the selected balls look like $AABCDEF$.  It can be arranged in $\frac{7!}{2!}=21$ ways. $AA$ can be $one$ out of $seven$. Given $AA$, $BCDEF$ has to be $five $ out of $six$. In total the probability is $\frac{7!\cdot 7\cdot 6}{2!\cdot 7^7}=\frac{3\cdot 6!}{7^5}\approx 12.85\%$
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