Q.7.19 - A First Course in Probability

Question

There are n items in a box labeled H and m in a box labeled T. A coin that comes up heads with probability p and tails with probability 1 − p is flipped. Each time it comes up heads, an item is removed from the H box, and each time it comes up tails, an item is removed from the T box. (If a box is empty and its outcome occurs, then no items are removed.) Find the expected number of coin flips needed for both boxes to become empty. Hint: Condition on the number of heads in the first n + m flips.

Step-by-Step Solution

Answer

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

Step 1: Given information

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 

Step 2: Solution

The solution will be shown below,

$E\left[X\right]=\sum _{i=1}^{n+m} E[X\mid Y=i]P\{Y=i\}$

$=\sum _{i=0}^{n+m} E[X\mid Y=i{]}^{n+m}{C}_i{p}^i(1-p{)}^{n+m-i}$

If the numeral of Heads in the first $\mathrm{n}+\mathrm{m}$ flips is $\mathrm{i},i\le n$, 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,

$\mathrm{E}\left[\mathrm{X}\right]=\sum _{i=0}^n {\left(\frac{n-i}{p}\right)}^{n+m}{C}_i{p}^i(1-p{)}^{n+m-i}+\sum _{i=n+1}^{n+m} {\left(\frac{i-n}{1-p}\right)}^{n+m}{C}_i{p}^i(1-p{)}^{n+m-i}$

Step 3: Final answer

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