Question

You have $3$ cabinets. Each of these three cabinets had two drawers. In one of the cabinets, there is a gold coin in one drawer and a silver in the other. In another there is a gold coin in both drawers. In the last cabinet there is a silver coin in each of the drawer. Now you walk into the room, not knowing the location of any if the coins, and pick a gold coin in one of the drawer. Now, what is the probability that the other drawer in the same cabinet contains a gold coin?

Answer 1

Let $E_1,E_2$ and $E_3$ be the events that cabinet $I,II$ and $III$ are chosen, respectively.

$\Rightarrow P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

Let ${}^{\prime }{E}^{\prime }$ be the event that the coin drawn is of gold.

$\Rightarrow P\left(\frac{E}{E_1}\right)=P$ (a gold coin from cabinet $I$) $=\frac{1}{2}=\frac{1}{2}$

$\Rightarrow P\left(\frac{E}{E_2}\right)=P$ (a gold coin from cabinet $II$) $=\frac{2}{2}=1$

$\Rightarrow P\left(\frac{E}{E_3}\right)=P$ (a gold coin from cabinet $III$) $=0.$

Now the probability that the other coib in the box is of gold $=$ the probability that gold coins is drawn from cabinet $II=P\left(\frac{E_2}{E}\right).$

$\Rightarrow P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right)P\left(\frac{E}{E_2}\right)}{P\left(E_1\right)P\left(\frac{E}{E_1}\right)+P\left(E_2\right)P\left(\frac{E}{E_2}\right)+P\left(E_3\right)P\left(\frac{E}{E_3}\right)}$

$=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{1}{2}\times \frac{1}{3}\times 1+\frac{1}{3}\times 0}$

$=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{6}}$

$=\frac{2}{3}.$

Hence, the answer is $\frac{2}{3}.$