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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: De铿乶e 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 2m.

(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?

Short Answer

Expert verified

According to the information's given,

a) E(X) =nm+1+j=1m1j

b) E(Y) =2m+nE(X)=2m+n(nm+1+j=1m1j)

Step by step solution

01

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].

02

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(Si=1)=1m+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=1if and only if that produced small pill has lasted longer than remaining n-j large pills. Therefore

P(Lj=1)=1nj+1

03

Explanation (part a)

Finally, we have

E(X)=i=1nE(Si)+j=1mE(Lj)=nm+1+j=1m1nj+1

=nm+1+j=1m1j

04

Final Answer (part a)

E(X) =nm+1+j=1m1j

05

Given Information (part b)

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

06

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(Y)=2m+nE(X)=2m+n(nm+1+j=1m1j)

07

Final Answer (part b)

E(Y)=2m+nE(X)=2m+n(nm+1+j=1m1j)

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