# What is the probability of picking $4$ balls out from the two bins?

Assume there are $2$ bins (bin $a$ and bin $b$ ), and each bin contains $n$ indistinguishable balls with $n>10$. The probability of picking $k$ ball from each bin is $p_k$, ($k\ge 0$). Then what is the probability of picking $4$ balls out from the two bins?

My try:

If $i$ balls are picked from bin $a$, then $4-i$ balls should be picked from bin $b$, with $0\le i\le 4$. Therefore, the probability of picking $4$ balls from the two bins is \begin{align*} &\Pr[\text{picking $4$ balls from the two bins}] \\ & = p_0\cdot p_4 + p_4\cdot p_0 \\ & + p_1\cdot p_3 + p_3\cdot p_1 \\ & + p_2\cdot p_2 \end{align*}

However, my classmate said my equation is not "symmetric", and he gave the probability as

\begin{align*} &\Pr[\text{picking $4$ balls from the two bins}] \\ & = p_0\cdot p_4 + p_4\cdot p_0 \\ & + p_1\cdot p_3 + p_3\cdot p_1 \\ & + p_2\cdot p_2 + \color{red}{ p_2\cdot p_2} \end{align*}

Which probability is right? If I am right, how do I convince him? Or did I miss something in my calculation?

• You probably meant to say: for each bin the probability that $k$ balls will be picked out of it is $p_k$. Commented Apr 4, 2014 at 9:55
Let $X$ be the number of balls picked from bin $a$, let $Y$ be the number of balls picked from bin $b$ and let $Z=X+Y$. Assume that $X$ and $Y$ are independent. The distribution of $Z$ is given by the discrete convolution formula $$\Pr\{Z=z\}=\sum_{x=0}^{z}\Pr\{X=x\}\Pr\{Y=z-x\}$$ for $0\le z\le 2n$, which is exactly what you obtained for $\Pr\{Z=4\}$.
You can convince your friend with a simple example. Suppose that $p_2=1$ and all other probabilities are $0$. Then the probability that your friend obtained is $2$, which is not very good...