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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Mon Feb 15, 2010 10:40 pm

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?
A) 1
B) 1/2
C) 2/3
D) 0
E) 1/2

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Source: Beat The GMAT — Problem Solving | Mon Feb 15, 2010 10:40 pm

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Thu Feb 18, 2010 4:51 am

shashank.ism wrote: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?
A) 1
B) 1/2
C) 2/3
D) 0
E) 1/2

Well,this question utilises Bayer's theorem and thus would be likely a "Challenge Problem".
It is also called the inverse probability theorem.
It goes like this:-P(A|B) = [P(B|A)][P(A)]/P(B) .

Over here it will be:- P(A) and P(B) will be same (1/3)
So, the answer will be [spoiler]1/2.[/spoiler]

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Mom4MBA: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Fri Jan 22, 2010 1:10 pmFollowed by: 4 members; Location: New Jersey; Thanked: 13 times; GMAT Score:640

by Mom4MBA » Fri Feb 19, 2010 7:37 am

Solution for this is explained in this pdf, question number 8

https://www.davidson.edu/academic/econom ... stions.pdf

answer 2/3

Stay focused

shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Fri Feb 19, 2010 9:54 am

Mom4MBA wrote:Solution for this is explained in this pdf, question number 8

https://www.davidson.edu/academic/econom ... stions.pdf

answer 2/3

Solution is well explained in the .pdf link thanks mom forMBA though I generally solve it by Baye's theorem as also suggested by harsh . champ.
baye's theorem makes the problem simpler and easily understandable.

Mom4MBA: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Fri Jan 22, 2010 1:10 pmFollowed by: 4 members; Location: New Jersey; Thanked: 13 times; GMAT Score:640

by Mom4MBA » Fri Feb 19, 2010 12:49 pm

Is this the way to solve using Baye's theorem,

P(A/G) = [P(G/A) P(A)] / P(G)

P(G/A) = 1
P(A) = 1/3
P(G) = 3/6 = 1/2

P(A/G) = 2/3

Stay focused

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Fri Feb 19, 2010 12:54 pm

Mom4MBA wrote:Is this the way to solve using Baye's theorem,

P(A/G) = [P(G/A) P(A)] / P(G)

P(G/A) = 1
P(A) = 1/3
P(G) = 3/6 = 1/2

P(A/G) = 2/3

Yes,this is the correct approach for Bayer's Theorem.

For further queries ,you can refer to the link:-
https://en.wikipedia.org/wiki/Bayes%27_theorem

Here,the concept of conditional probability has been explained in detail.
Hope it helps.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

Mom4MBA: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Fri Jan 22, 2010 1:10 pmFollowed by: 4 members; Location: New Jersey; Thanked: 13 times; GMAT Score:640

by Mom4MBA » Fri Feb 19, 2010 1:40 pm

Thanks for the help Harsh

Stay focused

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