<s>[INST] Hiding 3 coins in a pie, and slicing the pie in 8 equal pieces - probability Okay so I have this question on my maths sheet and it's in the permutations section but I can't get my head round it (probably just being dumb) 
I know that $P(n,r) = \frac{n!}{(n-r)!}$ 
and that $C(n,r) = \frac{n!}{r!(n-r)!}$
And old lady bakes a pie for her grandson, puts three coins into the pie and slices it into 8 equal portions. What is the probability that the grandson will find 2 or more coins in his slice of the pie, assuming that each coin is equally likely to be anywhere within the pie? 
 [/INST] If an event has probability $p$, then the probability that it will occur exactly $k$ times in $n$ trials is 
$$\binom{n}{k}p^k(1 - p)^{n - k}$$
Since the pie has been cut into eight pieces of equal size, the probability that a particular coin is in that slice is $p = 1/8$.  The number of trials is three since we must check if each coin is in that slice.  Hence, the probability that exactly two of the three coins will be found in a particular slice is 
$$\binom{3}{2}\left(\frac{1}{8}\right)^2\left(1 - \frac{1}{8}\right)^1$$
The probability that all three of the coins will be found in the same slice is 
$$\binom{3}{3}\left(\frac{1}{8}\right)^3\left(1 - \frac{1}{8}\right)^0$$
Thus, the probability that at least two coins will be found in the same slice is 
\begin{align*}
\binom{3}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^1 + \binom{3}{3}\left(\frac{1}{8}\right)^3\left(\frac{7}{8}\right)^0 & = 3 \cdot \frac{1}{64} \cdot \frac{7}{8} + 1 \cdot \frac{1}{512} \cdot 1\\ 
& = \frac{21}{512} + \frac{1}{512}\\ 
& = \frac{22}{512}\\
& = \frac{11}{256}
\end{align*}
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