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Hint: - This question may be interpreted in two ways, which we shall discuss separately.

I) If we consider that all numbers of shillings are equally likely, we shall have three hypothesis:

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${p_1} = 1,{p_2} = \dfrac{1}{2},{p_3} = \dfrac{1}{6}$

Hence probability of first hypothesis is${Q_1}$

$ = 1 \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{6}{{10}}$

Probability of second hypothesis is${Q_2}$

$ = \dfrac{1}{2} \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{3}{{10}}$

Probability of third hypothesis is${Q_3}$

$ = \dfrac{1}{6} \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{1}{{10}}$

Therefore the probability that another drawing will give a sovereign is

$

= \left( {{Q_1} \times 0} \right) + \left( {{Q_2} \times \dfrac{1}{4}} \right) + \left( {{Q_3} \times \dfrac{2}{4}} \right) \\

= \left( {\dfrac{6}{{10}} \times 0} \right) + \left( {\dfrac{3}{{10}} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{{10}} \times \dfrac{2}{4}} \right) \\

= \dfrac{3}{{40}} + \dfrac{2}{{40}} \\

= \dfrac{5}{{40}} = \dfrac{1}{8} \\

$

II) If each coin is equally likely to be a shilling or sovereign, by taking the terms in the expansion of

${\left( {\dfrac{1}{2} + \dfrac{1}{2}} \right)^4}$, we see that the chance of four shillings is$\dfrac{1}{{16}}$, of three shillings is$\dfrac{4}{{16}}$, of two shillings is$\dfrac{6}{{16}}$; thus

${P_1} = \dfrac{1}{{16}},{P_2} = \dfrac{4}{{16}},{P_3} = \dfrac{6}{{16}}$

Also, as before ${p_1} = 1,{p_2} = \dfrac{1}{2},{p_3} = \dfrac{1}{6}$.

Hence$\dfrac{{{Q_1}}}{6} = \dfrac{{{Q_2}}}{{12}} = \dfrac{{{Q_3}}}{6} = \dfrac{{{Q_1} + {Q_2} + {Q_3}}}{6} = \dfrac{1}{{24}}$

Therefore the probability that another drawing will give a sovereign

$

= \left( {{Q_1} \times 0} \right) + \left( {{Q_2} \times \dfrac{1}{4}} \right) + \left( {{Q_3} \times \dfrac{2}{4}} \right) \\

= \left( {\dfrac{1}{4} \times 0} \right) + \left( {\dfrac{1}{2} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{4} \times \dfrac{2}{4}} \right) \\

= 0 + \dfrac{1}{8} + \dfrac{2}{{16}} \\

= \dfrac{1}{4} \\

$

Note: - Both the methods used above are equally correct till the direction is not mentioned in the question. In case of mutually exclusive events, such as in the above probability of different events are found out separately and then added to find the final probability.

I) If we consider that all numbers of shillings are equally likely, we shall have three hypothesis:

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${p_1} = 1,{p_2} = \dfrac{1}{2},{p_3} = \dfrac{1}{6}$

Hence probability of first hypothesis is${Q_1}$

$ = 1 \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{6}{{10}}$

Probability of second hypothesis is${Q_2}$

$ = \dfrac{1}{2} \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{3}{{10}}$

Probability of third hypothesis is${Q_3}$

$ = \dfrac{1}{6} \div \left( {1 + \dfrac{1}{2} + \dfrac{1}{6}} \right) = \dfrac{1}{{10}}$

Therefore the probability that another drawing will give a sovereign is

$

= \left( {{Q_1} \times 0} \right) + \left( {{Q_2} \times \dfrac{1}{4}} \right) + \left( {{Q_3} \times \dfrac{2}{4}} \right) \\

= \left( {\dfrac{6}{{10}} \times 0} \right) + \left( {\dfrac{3}{{10}} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{{10}} \times \dfrac{2}{4}} \right) \\

= \dfrac{3}{{40}} + \dfrac{2}{{40}} \\

= \dfrac{5}{{40}} = \dfrac{1}{8} \\

$

II) If each coin is equally likely to be a shilling or sovereign, by taking the terms in the expansion of

${\left( {\dfrac{1}{2} + \dfrac{1}{2}} \right)^4}$, we see that the chance of four shillings is$\dfrac{1}{{16}}$, of three shillings is$\dfrac{4}{{16}}$, of two shillings is$\dfrac{6}{{16}}$; thus

${P_1} = \dfrac{1}{{16}},{P_2} = \dfrac{4}{{16}},{P_3} = \dfrac{6}{{16}}$

Also, as before ${p_1} = 1,{p_2} = \dfrac{1}{2},{p_3} = \dfrac{1}{6}$.

Hence$\dfrac{{{Q_1}}}{6} = \dfrac{{{Q_2}}}{{12}} = \dfrac{{{Q_3}}}{6} = \dfrac{{{Q_1} + {Q_2} + {Q_3}}}{6} = \dfrac{1}{{24}}$

Therefore the probability that another drawing will give a sovereign

$

= \left( {{Q_1} \times 0} \right) + \left( {{Q_2} \times \dfrac{1}{4}} \right) + \left( {{Q_3} \times \dfrac{2}{4}} \right) \\

= \left( {\dfrac{1}{4} \times 0} \right) + \left( {\dfrac{1}{2} \times \dfrac{1}{4}} \right) + \left( {\dfrac{1}{4} \times \dfrac{2}{4}} \right) \\

= 0 + \dfrac{1}{8} + \dfrac{2}{{16}} \\

= \dfrac{1}{4} \\

$

Note: - Both the methods used above are equally correct till the direction is not mentioned in the question. In case of mutually exclusive events, such as in the above probability of different events are found out separately and then added to find the final probability.