A purse contains five coins, each of which may be a shilling or a sixpence; two are drawn and found to be shillings: find the probable value of the remaining coins.

Hint: Here, in the given question, we have been given five coins in a purse each of which may either be a shilling or a six pence coin. Two coins were drawn and observed to be shillings and we are asked to find the total expected (probable) value of the remaining coins. We will take different cases of three coins drawn and then multiply the amount to their probabilities and then add them all to get the desired result.

Complete step-by-step answer:
Total number of coins - \[5\]
But two of them are already drawn, so we have to consider the three coins left in a purse.
Let \[A\] be the number of shilling coins drawn. So, \[A\] can take the value \[0,1,2,3\] and \[X\] denote the value(in shillings) of three coins withdrawn.
When three coins are drawn, we get
Sample space, \[S = \left\{ {SSS,SSP,SPS,SPP,PSS,PSP,PPS,PPP} \right\}\] where \[S,P\] denotes that the coin withdrawn is a shilling or a six pence respectively.
Case1. Three coins drawn are all sixpence
 \[
  P(A = 0) = P(PPP) \\
   \Rightarrow P(A = 0) = \dfrac{1}{8} \;
 \]
Case2. Out of three coin draws, one is shilling.
 \[
  P(A = 1) = P(SPP,PSP,PPS) \\
   \Rightarrow P(A = 1) = \dfrac{3}{8} \;
 \]
Case2. Out of three coin draws, two are shillings.
 \[
  P(A = 2) = P(SSP,SPS,PSS) \\
   \Rightarrow P(A = 2) = \dfrac{3}{8} \;
 \]
Case1. Three coins drawn are all shillings.
 \[
  P(A = 3) = P(SSS) \\
   \Rightarrow P(A = 3) = \dfrac{1}{8} \;
 \]
Thus, the probability distribution of \[X\] is:



Expected value of \[X\] , \[E\left( X \right) = \sum {X.P\left( X \right)} \]
Total probable value of the three coins = \[\left( {\dfrac{3}{2} \times \dfrac{1}{8}} \right) + \left( {2 \times \dfrac{3}{8}} \right) + \left( {\dfrac{5}{2} \times \dfrac{3}{8}} \right) + \left( {3 \times \dfrac{1}{8}} \right)\]
 \[ = 2.25\] shillings.
So, the correct answer is “ \[ 2.25\] shillings.”.

Note: We have taken \[A\] as the number of shillings drawn out of three coins and \[X\] as their corresponding amount (in shillings). Therefore, \[P(A) = P(X)\] for all corresponding values of \[A\] and \[X\] . In other words, if \[P\left( {A = 0} \right) = \dfrac{1}{8}\] , it means there are all six pence coins and the value of \[X\] becomes \[\dfrac{3}{2}\] or \[1.5\] shillings. Hence, we can write \[P\left( {A = 0} \right) = P\left( {X = \dfrac{3}{2}} \right) = \dfrac{1}{8}\] .