Box

[shashank.ism]

A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

1
1/2
2/3
0
1/2

[Ian Stewart]

A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

1
1/2
2/3
0
1/2

You could work this out from the answer choices by pure logic. Certainly there is some chance the coin comes from the drawer with two gold coins, so 0 is not the answer. Certainly there is some chance the coin comes from the drawer with one gold and one silver coin, so 1 is not the answer. And since only one answer can be correct on the GMAT, neither of the answers which read '1/2' can be correct (then there would be two different answer choices, both of which are correct), so the answer must be 2/3.

Alternatively: we know in advance that we have picked a gold coin. There are 3 gold coins we could have picked, and 2 of them come from the drawer with two gold coins. So the probability our coin comes from that drawer is 2/3. You could also use a conditional probability formula here.

[ldoolitt]

I know that they don't test Bayes Rule on the test but it would make this problem super handy.

Bayes Rule

P(A | B) = P(B | A) * P(A) / P(B)

In this case you have

P(You picked from 2 gold coin drawer GIVEN you selected a gold coin) =

P(You picked a gold coin GIVEN you are picking from the drawer with 2 gold coins) *
P(You picked the drawer with 2 coins) /
P(You picked a gold coin)

P(You picked a gold coin GIVEN you are picking from the drawer with 2 gold coins) = 1 (obviously, you have to pick a gold coin!)

P(You picked the drawer with 2 coins) = 1/3 because there are 3 drawers
P(You picked a gold coin) = 1/2 because half of the total coins are gold

Therefore

P(You picked from 2 gold coin drawer GIVEN you selected a gold coin) =
1 * 1/3 / 1/2 = 2/3

[shashank.ism]

A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

1
1/2
2/3
0
1/2

You could work this out from the answer choices by pure logic. Certainly there is some chance the coin comes from the drawer with two gold coins, so 0 is not the answer. Certainly there is some chance the coin comes from the drawer with one gold and one silver coin, so 1 is not the answer. And since only one answer can be correct on the GMAT, neither of the answers which read '1/2' can be correct (then there would be two different answer choices, both of which are correct), so the answer must be 2/3.

Alternatively: we know in advance that we have picked a gold coin. There are 3 gold coins we could have picked, and 2 of them come from the drawer with two gold coins. So the probability our coin comes from that drawer is 2/3. You could also use a conditional probability formula here.

Hmm alternative seems to be very good and it helps me to think this way.
Otherwise I could have gone for baye's theorem to solve the problem..Though it also solves the problem but a bit of calculation is needed ..

[ldoolitt]

Hmm alternative seems to be very good and it helps me to think this way.
Otherwise I could have gone for baye's theorem to solve the problem..Though it also solves the problem but a bit of calculation is needed ..

Agreed, always just trying to provide a different approach!

[harsh.champ]

A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

1
1/2
2/3
0
1/2

I think this is an advanced probability question involving concepts like bayer's rule so as such would not feature in the GMAT.
Anyways we can rule out D.
By logic,P(2 gold coins)=3C1 X 1C1 = 3
P(total) = 3C1 x 1C1 + 3C1 x 2C1 = 3 + 6 =9
Hence,[spoiler]C. 2/3[/spoiler]

Is this approach incorrect??

[komal]

A box has three drawers; one contains two gold coins, one contains two silver coins, and one contains one gold coin and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is one that contains two gold coins?

1
1/2
2/3
0
1/2

Here's another approach that i found interesting :