Question

In a purse there are 10 coins, all shillings except one which is a sovereign; in another, there are 10 coins all shillings. 9 coins are taken from the former purse and put into the latter, and the 9 coins are taken from the latter and put into the former. Find the chance that the sovereign is still in the first purse.

Answer 1

Hint: Here we will consider the probabilities of moving coins from one purse to another to find the chance that the sovereign is still in the first purse.

Complete step-by-step answer:

Now in the problem it is given that we have 2 purses.
In purse1- 10 total coins out of which 9 shilling and 1 sovereign.
In purse2- 10 total coins all are shilling.
Now we have taken 9 coins from purse 1 and put into purse 2 and then 9 coins are taken from purse2 and put into purse 1.
Now previously purse 1 has 1 sovereign coin so we need to tell the probability that the sovereign coin is still in the first purse. Now let’s first talk about the probability that sovereign is in purse 2, there can be two conditions for this
${E_1} = $The sovereign was amongst the 9 coins moved to the second purse
${E_2} = $ The sovereign was not among the 9 coins moved again to the first purse.
Now $P({E_1}) = \dfrac{9}{{10}}$i.e. $\dfrac{{{\text{favorable}}}}{{total}}$
$P({E_2}) = \dfrac{{10}}{{19}}$ Using the same above concept.
Now, $P({E_1} \cap {E_2}) = P({E_1}) \times P({E_2}) = \dfrac{9}{{10}} \times \dfrac{{10}}{{19}} = \dfrac{9}{{19}}$
Hence this was probability that sovereign is in purse 2 so the required probability that sovereign is in purse one is$1 - \dfrac{9}{{19}} = \dfrac{{10}}{{19}}$.
Hence the required answer is $\dfrac{{10}}{{19}}$.

Note: In such probability questions you can have an approach like thinking of all the converse cases of the required event and then subtracting it with 1 because we know $P(E) = 1 - \overline {P(E)} $