Q.7.24 - A First Course in Probability

Question

A bottle initially contains m large pills and n small pills. Each day, a patient randomly chooses one of the pills. If a small pill is chosen, then that pill is eaten. If a large pill is chosen, then the pill is broken in two; one part is returned to the bottle (and is now considered a small pill) and the other part is then eaten.

 (a) Let X denote the number of small pills in the bottle after the last large pill has been chosen and its smaller half returned. Find E[X].

 Hint: Define n + m indicator variables, one for each of the small pills initially present and one for each of the small pills created when a large one is split in two. Now use the argument of Example $2$m.

(b) Let Y denote the day on which the last large pills chosen. Find E[Y]. 

Hint: What is the relationship between X and Y? 

Step-by-Step Solution

Answer

According to the information's given,

a) E(X) =$\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j}$

b) E(Y) = $2m+n-E\left(X\right)=2m+n-(\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j})$

Step 1: Given Information (part a)

Let X denote the number of small pills in the bottle after the last large pill has been chosen and its smaller half returned. Find E[X]. 

Step 2: Explanation (part a)

Define indicator random variables Si that are equal to 1 if and only if i th small pill has survived until all the large pills have been divided. Similarly, define indicator random variable Lj that marks if small pill produced from j th large pill that has been divided and eaten (in the row) has survived until the end. Observe that

$P(S_i=1)=\frac{1}{m+1}$

since we consider m large pills and that certain small pill, and because of the symmetry of the problem we have that probability. On the other hand, Lj=$1$ if and only if that produced small pill has lasted longer than remaining n-j large pills. Therefore

$P(L_j=1)=\frac{1}{n-j+1}$

Step 3: Explanation (part a)

Finally, we have

$E\left(X\right)=\sum _{i=1}^{n}E\left(S_i\right)+\sum _{j=1}^{m}E\left(L_j\right)=\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{n-j+1}$

$=\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j}$

Step 4: Final Answer (part a)

E(X) =$\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j}$

Step 5: Given Information (part b)

Let Y denote the day on which the last large pills chosen. Find E[Y].  

Step 6: Explanation (part b)

Observe that all the pills will be eaten in 2m+n days. If there remained X small pills in the bottle, that means that we have been eating for 2 m+n-X days which is Y. Therefore

$E\left(Y\right)=2m+n-E\left(X\right)=2m+n-(\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j})$

Step 7: Final Answer (part b)

$E\left(Y\right)=2m+n-E\left(X\right)=2m+n-(\frac{n}{m+1}+\sum _{j=1}^m\frac{1}{j})$