Question

There are three boxes each having two drawers. The first box contains a gold coin in each drawer. The second box contains a gold coin in one drawer and a silver coin in another. The third box contains a silver coin in each drawer. A box is chosen at random and a drawer is opened. If the gold coin is found in that drawer, then the probability that the other drawer also contains a gold coin is?
A) \[\dfrac{1}{3}\]
B) \[\dfrac{2}{5}\]
C) \[\dfrac{2}{3}\]
D) \[\dfrac{3}{5}\]

Answer 1

Hint:
Here we will find the probability of choosing a box as well as the probability of finding the gold coin. We will use Bayes theorem of probability which is used when the probability of occurring event is based on some prior event.

Complete step by step solution:
Given that there are three boxes each having two drawers.
 Let the boxes be \[{b_1},{b_2},{b_3}\].
Now the probability of selecting a box at random is
\[\dfrac{{number{\text{ }}of{\text{ }}box{\text{ }}to{\text{ }}be{\text{ }}selected}}{{total{\text{ }}number{\text{ }}of{\text{ }}boxes}} = \dfrac{1}{3}\].
But the process is not completed yet. There are two drawers to each box with some mystery! Such that,
1) The first box contains a gold coin in each drawer.
2) The second box contains a gold coin in one drawer and a silver coin in another.
3) The third box contains a silver coin in each drawer.
Now we are given a condition that a box is chosen at random and a drawer is opened. If the gold coin is found in that drawer, then the probability that the other drawer also contains a gold coin!
Now we have to find the probability of gold coin in the drawer of the box
For \[{b_1}\] it is \[1\] because both drawers have gold coins.
For \[{b_2}\] it is \[\dfrac{1}{2}\] because that box is having one gold coin in any one of the drawers.
For \[{b_3}\] it is zero because it contains all two silver coins. No gold coin.
Let’s find the probability using Bayes theorem,
\[p\left( {A\left| B \right.} \right) = \dfrac{{p\left( {B\left| A \right.} \right).p(A)}}{{p(B)}}\]
Where \[p\left( {A\left| B \right.} \right)\] is the probability of A given that B is true. And \[p\left( {B\left| A \right.} \right)\] is the probability of B given that a is true.
So let’s put the respective probabilities
For the event here we have to find the probability of a gold coin in the drawer of the box, so let A is the event of finding a gold coin and B is the event of box selection.
\[ \Rightarrow \dfrac{{1 \times \dfrac{1}{3}}}{{1 \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{1}{3} + 0 \times \dfrac{1}{3}}}\]
\[ \Rightarrow \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3} + \dfrac{1}{6}}}\]
Taking LCM in denominator,
\[ \Rightarrow \dfrac{{\dfrac{1}{3}}}{{\dfrac{{2 + 1}}{6}}}\]
\[\begin{gathered}
   \Rightarrow \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{2}}} \\
   \Rightarrow \dfrac{2}{3} \\
\end{gathered} \]

Hence the correct option is C.

Note:
Here don’t just find the probability as only one box is having two gold coins so it should be \[\dfrac{1}{3}\] only. No because first we will select the box and then the gold coin condition.