There are 3 bags. One contains 2 gold coins, another has

2 silver coins and the 3rd has one of each. You pick a bag

at random, and without looking inside take out one coin.

It's gold. What is the probability that the other coin

in that bag is also gold?

## 3 comments:

the answer is .5

bcoz the coin taken out is gold(the bag containing both silver is not to be taken into account)

thn there r only 2 case are possible either a silver coin or a gold coin

@Sajal

Your answer results from the commonly-made assuption that the two possible cases are equally likely. However, if you add up the possible complete probabilities using that assumption, they don't sum to 1.

(Chance of selecting any bag = 1/3.

If after selecting a gold coin the prob'y of it being either of the possible cases is 1/2;

then each of those has a complete prob of 1/2 * 1/3 = 1/6 for a total of 1/3 if a gold coin is drawn.

It's the same if the coin you draw is silver - total prob'y is 1/3. These only sum to 2/3).

One way to look at this problem is that there are 6 equally likely ways to select a coin. These are:

Bag CoinG+G G

G+G G

G+S G

G+S S

S+S S

S+S S

All you have done by identifying the drawn coin as gold is eliminate the last 3 cases. It can be seen from the remaining cases that it is twice as likely that you initially selected the G+G bag.

So the prob'y of the other coin being gold is 2/3.

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btw, thanks for turning HTML and preview on, Eldhose. :-)

It is 1/2.

He took a Gold coin already. It means he eliminated the Silver-Silver bag.

That 1 Gold would have come from Gold-Gold bag or Gold-Silver bag.

He took a Gold, it means that would have come from one of those two bags.

The other coin's possibilities are Gold from Gold-Gold bag or Silver from Gold-Silver bag.

Hence 1/2.

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