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A coin that comes up heads with probability p and tails with probability 1 − p is flipped.

It comes up heads, an item is removed from the H box, and each time it comes up tails

Item is removed from the T box

The solution will be shown below,

$E\left[X\right]=\underset{i=1}{\overset{n+m}{∑}} E[Xâˆ\pounds Y=i]P\{Y=i\}$

$=\underset{i=0}{\overset{n+m}{∑}} E[Xâˆ\pounds Y=i{]}^{n+m}{C}_i{p}^i(1−p{)}^{n+m−i}$

If the numeral of Heads in the first $\mathrm{n}+\mathrm{m}$flips is $\mathrm{i},i≤n$, Then many more flips is the numeral of flips needed to receive additional (n-i) Heads. Similarly, if the number of Heads in the first $n+m$flips is $i,j>n$, then because there would keep existed a total of $n+m-i<m$tails, the numeral of more flips is the number required to receive an additional ($\mathrm{i}-\mathrm{n}$) Heads. Since the number of flips needed for j outcomes of a particular type is a negative Binomial Random variable whose mean is j divided by the probability of that outcome,

Therefore,

$\mathrm{E}\left[\mathrm{X}\right]=\underset{i=0}{\overset{n}{∑}} {\left(\frac{n−i}{p}\right)}^{n+m}{C}_i{p}^i(1−p{)}^{n+m−i}+\underset{i=n+1}{\overset{n+m}{∑}} {\left(\frac{i−n}{1−p}\right)}^{n+m}{C}_i{p}^i(1−p{)}^{n+m−i}$

The expected number of coin flips needed for both boxes to become empty is $\mathrm{E}\left[\mathrm{X}\right]=\underset{i=0}{\overset{n}{∑}} {\left(\frac{n−i}{p}\right)}^{n+m}{C}_i{p}^i(1−p{)}^{n+m−i}+\underset{i=n+1}{\overset{n+m}{∑}} {\left(\frac{i−n}{1−p}\right)}^{n+m}{C}_i{p}^i(1−p{)}^{n+m−i}$.