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You are not logged in. Pages: 1 #1 20130125 08:45:07
can i solve conditional probability using n(E)/n(S) approachQ. there are two bags A and B; bag A contains 2 white and 3 black balls; bag B contains 3 white and 4 black balls. A bag is chosen at random find the probability that the ball drawn is white. #2 20130125 10:27:03
Re: can i solve conditional probability using n(E)/n(S) approachhi ashmath, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #4 20130127 19:31:32
Re: can i solve conditional probability using n(E)/n(S) approachhi ashmath So, how to work out the total number of outcomes? My approach is to imagine the 'experiment' is repeated again and again until every event has had a fair chance to occur. The key here is what do I mean by a fair chance. For bag A there are 5 balls so I can get the probabilities just by saying 'total number of outcomes' = 5. For bag B in the same way 'total number of outcomes' = 7 So how to fairly decide on the number when either bag may be picked. If I pick from bag A 35 times (5x7) then P(W) = 14/35 and P(B) = 21/35 and for bag B, I would get P(W) = 15/35 and P(B) = 20/35 So I can treat A and B fairly by having 35 trials for A and 35 trials for B. So I'll have 70 trials altogether. 70 splits like this P(A) = 35/70 and P(B) = 35/70 Then use the probabilities above to get P(A & W) = 14/70 P(A & B) = 21/35 P(B & W) = 15/70 P(B & B) = 20/35 Then I can see that 'total number of outcomes' = 70 'Number of ways W happens' = 14 + 15 = 29 So P(W) = 29/70 Hope that helps, Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei Pages: 1 