Solve the following:

Given three identical boxes, I, II, and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

#### Solution

Let event B_{1}: Selecting box I have two gold coins,

event B_{2}: Selecting box II having two silver coins,

event B_{3}: Selecting box III having one silver and one gold coin,

event G: Coin is gold.

P(B_{1}) = P(B_{2}) = P(B_{3}) = `1/3`

`"P"("G"/"B"_1) = 1, "P"("G"/"B"_2) = 0, "P"("G"/"B"_3) = 1/2`

P(G) = `"P"("B"_1) "P"("G"/"B"_1) + "P"("B"_2) "P"("G"/"B"_2) + "P"("B"_3) "P"("G"/"B"_3)`

= `1/3[1 + 0 + 1/2]`

= `1/3(3/2)`

= `1/2`

To find the probability that the other can in the box is also gold. Which is possible only when it is drawn from box I.

∴ Required probability = `"P"("B"_1/"G")`

By Bayes’ theorem,

`"P"("B"_1/"G") = ("P"("B"_1)"P"("G"/"B"_1))/("P"("G"))`

= `(1/3(1))/(1/2)`

= `2/3`