Question

A purse contains $4$ coins which are either sovereigns or shillings; $2$ coins are drawn and found to be shillings: if these are replaced what is the chance that another drawing will give a sovereign?

Answer 1

This question may be interpreted in two ways, which we shall discuss separately.
I. If we consider that all numbers of shillings are a priori equally likely, we shall have three hypotheses; for (i) all the coins may be shillings, (ii) three of them may be shillings, (iii) only two of them may be shillings.
Here $P_1=P_2=P_3;$
also $p_1=1,p_2=\frac{1}{2},p_3=\frac{1}{6}$.
Hence probability of first hypothesis$=1÷\left(1+\frac{1}{2}+\frac{1}{6}\right)=\frac{6}{10}=Q_1,$
probability of second hypothesis$=\frac{1}{2}÷\left(1+\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{10}=Q_2$,
probability of third hypothesis$=\frac{1}{6}÷\left(1+\frac{1}{2}+\frac{1}{6}\right)=\frac{1}{10}=Q_3$.
Therefore the probability that another drawing will give a sovereign$=(Q_1\times 0)+\left(Q_2\times \frac{1}{4}\right)+(Q_3\times \frac{2}{4})$
$=\frac{1}{4}\cdot \frac{3}{10}+\frac{2}{4}\cdot \frac{1}{10}=\frac{5}{40}=\frac{1}{8}$.
II. If each coin is equally likely to be a shilling or a sovereign, by taking the terms in the expansion of ${\left(\frac{1}{2}+\frac{1}{2}\right)}^4$, we see that the chance of four shillings is $\frac{1}{16}$, of three shillings is $\frac{4}{16}$, of two shillings is $\frac{6}{16};$ thus
$P_1=\frac{1}{16},P_2=\frac{4}{16},P_3=\frac{6}{16}$
also, as before, $p_1=1,p_2=\frac{1}{2},p_3=\frac{1}{6}$.
Hence $\frac{Q_1}{6}=\frac{Q_2}{12}=\frac{Q_3}{6}=\frac{Q_1+Q_2+Q_3}{24}=\frac{1}{24}$.
Therefore the probability that another drawing will give a sovereign$=(Q_1\times 0)+(Q_2\times \frac{1}{4})+(Q_3\times \frac{2}{4})$
$=\frac{1}{8}+\frac{2}{16}=\frac{1}{4}$.