# 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?

Toolbox:
• $P(E_1/A)=\large\frac{P(E_1)P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$
Step 1:
Let $E_1,E_2$ and $E_3$ be the event that boxes I,II and III are chosen respectively.
The $P(E_1)=P(E_2)=P(E_3)=\large\frac{1}{3}$
Let the event A be the event that the coin drawn in of gold.
Then $P(A/E_1)$=P(a gold coin from bag I)=$\large\frac{2}{2}$=1
$P(A/E_2)$=P(a gold coin from bag II)=0
$P(A/E_3)$=P(a gold coin from bag III)=$\large\frac{1}{2}$
Step 2:
The probability that the other coin in the box is of gold=the probability that gold coin is drawn from the box I
$\Rightarrow P(E_1/A)$
By Baye's theorem ,we know that
$P(E_1/A)=\large\frac{P(E_1)P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$
$\Rightarrow \large\frac{1/3\times 1}{1/3\times 1+1/3\times 0+1/3\times 1/2}$
$\Rightarrow \large\frac{2}{3}$
Hence the required probability is $\large\frac{2}{3}$
sir i have a doubt.... As to check whether second coin is of gold, we must have first golden coin... gold coins are only present in 2 boxes among given 3... so if A be the event of getting first coin as golden p(A)=2/3...
then further for getting second golden coin we must have to select box I among Box I and II... so If B is event of getting second golden coin then p(B)=1/2...
please tell me what part is incorrect in my logic....
thanking you,