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A clarification required from STOLYAR [#permalink]
03 Oct 2003, 22:59

Hi stolyar
I needed some clarification on one of the questions u had put in the excel tool created by this site. Its on probability. It said that there are 6 commodities A B C D E F and we have to choose 3 . E has already been choosen and what is the probability of choosing in the remaining 2.

My solution. as E has already been chosen 2 commodities can be choosen out of 5 in 2C5 ways = 10 ways.
Also B can choosen in 4 different ways. BA, BC , BD, BF so the probability is 4/10 =.4

Your solution P(E^B) = 1/30 (1/5*1/5)
P(E)= 1/6
therefore P(B)= P(E^B) / P(E) = .2
Fair enuf I would say. But why isnt the same answer coming from my method. Probably am missing out something. I wpould appreciate ur comment.

this question is not mine, so I need some clarification:

Your verbiage says: there are 6 commodities A B C D E F and we have to choose 3 . E has already been choosen and what is the probability of choosing (WHAT? ANY?) in the remaining 2.

sorry stolyar... I just missed "B" in my last sentence. E has already been chosen and we need to chose 2 more from the remaining 5 . so what is the probablility that one of them would be B. And the excel had ur name above the question so I presumed it was urs.
thanks for ur prompt reply
neeraj

the verbiage is slightly ambiguous. there should be words: provided that E is already taken. This would be a question about conditional probability: to find P of smth if smth else has already happened.

P(E B any)=P(E)*P(B any/E)
P(B any/E)=P(E B any)/P(E)
P(E B any)=1/6*1/5*4/4=1/30
P(E)=1/6

P=1/30:1/6=1/5=0.2

Your approach is for the following question: there are 5 commodities A B C D F and we have to choose 2 . What is the probability of choosing B in the remaining 2?

thanks stolyar... but just one thing. Once E has been taken isnt our question reduced to 5 commodities from which we have to pick 2. So why cant we use the same logic as mine in that case.

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