I was busy practicing probability questions and formed this question as a derivation of another (perhaps famous?) example. I shall quickly include the original question (in the original question drawers were used, but I will describe it with bags)
There are 3 bags with each 2 coins. bag 1 has 2 gold coins, bag 2 has 2 silver coins, and bag 3 has 1 gold and 1 silver coin. Somebody picks a bag and then a coin at random and it turns out to be gold. What is the probability that the other coin in the bag is also gold. Answer: 2/3.
Now I know how to calculate that, but I was curious to know what would happen with the probability when we put an extra gold coin in bag 3, meaning bag 3 would have 2 gold and 1 silver coins.
I calculated P(G) as 1/3 + 1/3*2/3 = 5/9 and then questioned myself why this isn't equal to 4/7 (the amount of gold out of the total coins).
So I came up with the following question:
We have 4 gold and 3 silver coins in 1 bag, so P(G) = 4/7. We then randomly divide these coins over 3 bags, given that every bag needs to have at least 1 coin and subsequently let another person choose a bag at random and then 1 coin at random. What is then the probability that the person draws a gold coin? Is this equal to 4/7 or did the probability somehow change? If so, why?
And then why does the probability of drawing a gold coin change, when we put all 7 coins in three bags as (2G),(2S),(2G, 1S), even when all three bags have an equal probability of being chosen?