You, your parents, your sister, go to visit grandma for her birthday. Grandma made a cake for the party. If she puts $20$ raisins in the cake at random in the cake, and she divides the cake into $5$ equal pieces, what's the probability that your sister gets as many raisins as you?
Let $X$ be the number of raisins in your piece. Let $Y$ be the number of raisins in her piece.
Since there are not that many raisins, you can just go through all the possible numbers of raisins that both you and your sister would get : $P(X=Y)=\Sigma_{k=0}^{10}P(X=Y=k)=\Sigma_{k=0}^{10}\binom{20}{k}\binom{20-k}{k}0.2^{2k}0.6^{20-2k}$ which is about $0.1406$.
$P[S_n=k|\theta]=$ $ n \choose k$ $\theta^k (1-\theta)^{n-k}$
where $S_n=X_1+X_2+...+X_n$ and $S_n \sim Bin(n,\theta)$
1.I don't understand why
$P(X=Y)=\Sigma_{k=0}^{10}P(X=Y=k)=\Sigma_{k=0}^{10}\binom{20}{k}\binom{20-k}{k}0.2^{2k}0.6^{20-2k}$
Shouldn't it be
$P(X=Y)=\Sigma_{k=0}^{10}P(X=Y=k)=\Sigma_{k=0}^{10}\binom{20}{k}\binom{20-k}{k}0.2^{2k}0.8^{20-2k}$?
2.I also don't understand why we have a product of binomial coefficients
$\binom{20}{k}\binom{20-k}{k}$. rather than one binomial coefficient either $\binom{20}{k}$ or $\binom{20-k}{k}$.
3.I also don't understand why we have exponentiation to the power of $2k$ and $20-2k$ rather than $k$ and $20-k$.
Would someone explain?
There's an equal chance for each piece to have a given raisin. So each raisin has a $0.2$ chance to end up on your cake and likewise $0.2$ chance to end up on your sisters. This means that there's a $0.6$ chance it ends up on someone else's cake. So for the inside of the sum we will have $0.2^k (0.2)^k (0.6)^{20-k - k}$ which by algebra simplifies to $0.2^{2k} (0.6)^{20-2k}$
We have a product of binomials because we have to choose twice. Once for the number of ways to choose exactly $k$ raisins out of $20$ for your cake, and another to choose exactly $k$ raisins from the remaining $20 - k$ raisins for your sisters cake.
See my answer for number one, as this pretty much addresses this.