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? 
 A: Hint
Let $A_i$ be the event that coin $i$ is in the grandson's piece.
2 coins in the grandson's piece and the other not: $\mathbb{P}(A_1\land A_2\land\lnot A_3)+\mathbb{P}(A_1\land\lnot A_2\land A_3)+\mathbb{P}(\lnot A_1\land A_2\land A_3)$.
3 coins in the grandson's piece: $\mathbb{P}(A_1\land A_2\land A_3)$.
Each of these can be further split multiplicatively assuming the events are independent, which is a necessary assumption to solve the problem that was not given.
A: I first choose 2 coins from 3: there are 3 ways to do this. What is the probability that a coin is in a given slice? It is $\frac18$, since it is equally likely to be in any of the 8 slices. So, the probability that both the coins are in a given slice is $\frac18$ AND $\frac18$ which is $\frac{1}{64}$. Since we can choose 2 coins in 3 ways, as mentioned before, we get $\frac{3}{64}$. Note that we have not said anything about the third coin: it may or may not be in the same slice. But we have counted the case in which all three coins are present thrice, so we must subtract twice the probability of all three coins being in the same slice, which is $2\left(\frac18 \cdot \frac18 \cdot \frac18\right) = \frac{2}{512}$. This gives us $\frac{3}{64} - \frac{2}{512} = \frac{22}{512}$.
A: 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*}
