All That Glitters Is Gold

We can throw out the box with silver coins. Label the gold coins in box $1$ as $G_{1A}$ and $G_{1B}$ and the gold coin in box $3$ as $G3.$ You have picked one of those coins at random; the probability that you picked a specific gold coin is equal for each of the coins. In $2$ of the cases, $G_{1A}$ and $G_{1B},$ the other coin $(G_{1B}$ and $G_{1A}$ respectively) is gold. Only if you pick $G_3$ would the other coin in the box be silver. Thus, the probability that the other coin is gold is equal to $\frac{2}{3},$ so $A=2$ and $B=3$ and $A+B=\boxed{5}$

I did this by drawing a tree diagram.

Box 1 must be G1 and G2

Box 2 must be S1 and S2

Box3 can be G1 and S1; or S1 and G1

$P(G2|G1) = \frac{P(G1and G2)}{P(G1)} =\frac{\frac{1}{3}}{\frac{1}{2}} =\frac{2}{3}$

$a+b=2+3=5$

Obviously, it's twice as likely that you'll have picked the box with $2$ gold coins, so the probability is that it's the one with $2$ gold coins is $\frac{2x}{x+2x} \Longrightarrow \frac{2}{3} \Longrightarrow \boxed{5}$ .

The question is deceptive, in the good way! There is only one box that contains 2 gold coins. So what we're really after is: given that we just picked a gold coin, what is the probability we picked the box with 2 gold coins? Plugging this into Bayes theorem quickly produces the answer.

The box containing 2 silver coins is of no use. Leaving that we have 4 coins out of which 3 are gold and 1 is silver. Since the picked one is a gold coin we are left with 3 coins out of which 2 are gold. Thus the probability is 2 out of 3.

use bayes theorem to find the probability of the coin coming from the box having 2 gold coins that comes out to be 2/3 then, since and b are coprime, a= 2 , b= 3, therefore a+b = 5

I did it using Bayes' theorem

we see the second box dosent contai any gold coin so we neglect it ... now probability of selecting first box from three boxes is 1/3 and if e draw one coin then probability of other coin be gold =1/1 so total probability =1/3 1/1 same case goes with 3box where we have 1/3 1+1/3*0 zero case suggest gold coin has been picked up and remaining is silver so probability of gold is zero so total probability =1/3+1/3=2/3 a=2 b=3 so a+b=5