Solution: single-dose tablets

This is the solution to the divertimento Single-dose tablets. 

The fun

A bottle contains 50 single-dose tablets of a medicine and 50 double-dose tablets. On arrival, each patient randomly takes out a pill; if it is a single-dose pill, he takes it, and if not, he divides it with a pill cutter, takes half of it and returns the other half as a new single-dose pill to the bottle. When five patients have passed,

1. What possible compositions of single- or double-dose pills can the bottle have?
2. What is the probability that the bottle has 47 single-dose pills?

Solution

When a single-dose tablet is removed, the number of double tablets remains the same and the number of single-dose tablets decreases by one; when a double-dose tablet is removed, the number of double-dose tablets decreases by one and the number of single-dose tablets increases by one. Thus, if out of the five patients there are \(x\) who take double pills and \(5-x\) who choose single-dose pills, the number of double pills will be \(D=50-x\), while the number of single-dose pills will be \(S=50+x-(5-x)=45+2x\).

Since \(x\) can take the values \(0, 1,\ldots,5\), the possible compositions \((D,S)\) will be

$$ (50,45),\;\;(49,47), \;\; (48,49), \;\; (47,51), \;\; (46,53), \;\; (45,55).$$

To have \(S=47\), it must be the case that \(x=1\), i.e., out of the five patients only one must take a double pill. Thus the possibilities are

$$ DSSSS, \;\; SDSSS, \;\; SSDSS, \;\; SSSDS, \;\; SSSSD.$$

The probabilities of each of these events are:

$$ P(DSSSS)=\frac{50}{100}\frac{51}{100}\frac{50}{99}\frac{49}{98}\frac{48}{97}\simeq 0,\!031865$$

$$ P(SDSSS)=\frac{50}{100}\frac{50}{99}\frac{50}{99}\frac{49}{98}\frac{48}{97}\simeq 0,\!031556$$

$$ P(SSDSS)=\frac{50}{100}\frac{49}{99}\frac{50}{98}\frac{49}{98}\frac{48}{97}\simeq 0,\!031240$$

$$ P(SSSDS)=\frac{50}{100}\frac{49}{99}\frac{48}{98}\frac{50}{97}\frac{48}{97}\simeq 0,\!030918$$

$$ P(SSSSD)=\frac{50}{100}\frac{49}{99}\frac{48}{98}\frac{47}{97}\frac{50}{96}\simeq 0,\!030589$$

In conclusion, the probability that there are 47 single-dose tablets (and necessarily 49 double-dose tablets) is approximately \(0,\!15617\).