Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
A clarification required from STOLYAR [#permalink]
03 Oct 2003, 22:59
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)
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.
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
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
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?