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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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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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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
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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.
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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
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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. :)
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by Mom4MBA » Fri Feb 19, 2010 1:40 pm
Thanks for the help Harsh :)
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