Question

One purse contains 5 shillings and 1 sovereign: a second purse contains 6 shillings. Two coins are taken from the first and placed in the second: then 2 are taken from the second and placed in the first: find the probable value of the contents of each purse.

Answer 1

Hint: In order to find the probable value of contents of each purse, we find the probabilities of individual events, using it we find the probability value of one of the contents, subtract it from one to find the other.

Complete step-by-step answer:
Given Data,
First purse: 5 shillings, 1 sovereign
Second purse: 6 shillings

We know the probability of an event is given by the formula,\[{\text{Probability = }}\dfrac{{{\text{favorable chances}}}}{{{\text{total number of chances}}}}\]

The chance that the sovereign is in the first purse is equal to the sum of chances that it has moved twice and that it has not moved at all.
That is, the chance = $\dfrac{1}{3}.\dfrac{1}{4} + \dfrac{2}{3}.1 = \dfrac{3}{4}$

Also we know that the probability of an even not occurring can be expressed as,
P(not occurring) = 1 – P(event occurring)
As we know that the sum of probabilities of an event is always equal to 1.

∴the chance that the sovereign is in the second purse = 1 - $\dfrac{3}{4} = \dfrac{1}{4}$
Hence the probable value of first purse = $\dfrac{3}{4}{\text{ of 25s + }}\dfrac{1}{4}{\text{ of 6s = 1}}{\text{.0s}}{\text{.3d}}$
∴the probable value of second purse = 31s.20$\dfrac{1}{4}$s = 10s.9d

Or the problem may be solved as follows
The probable value of the coins removed = $\dfrac{1}{3}$ of 25s = 8$\dfrac{1}{3}$s
The probable value of the coins bought back = $\dfrac{1}{4}{\text{ of }}\left( {6{\text{s + 8}}\dfrac{1}{3}{\text{s}}} \right) = 3\dfrac{7}{{12}}{\text{s}}$
Therefore, probable value of first purse = $\left( {25 - 8\dfrac{1}{3} + 3\dfrac{7}{{12}}} \right){\text{ shillings = 1}}{\text{.0s}}{\text{.3d}}$

Note: In order to solve questions of this type the key is to have a good understanding of probability concepts is required. For two events, probabilities of both of them added, is always equal to one. While calculating the probability of an event we multiply the possibilities that occur with each other and add the events that occur in a serial order.