HI Experts,
Here is a question I am getting confused with (Source: 8th Grade maths):
A bag of marbles contains 12 blue marbles and 1 white marble. If you reach into the bag without looking, what is the probability of selecting the white marble on the first or second draw?
Since it is question picked up not from a GMAT book it doesn't have options.
I solve it like this:
P( white on first) = 1/13.
P(white on second) = P(blue on first) * P(white on second) = 12/13 * 1/12 = 1/13
Thus answer P( white on first or white on second) = 1/13 + 1/13 = 2/13.
However, the book solves the question in this way:
P(white on first) = 1/13 (Matches with my answer so far)
P(white on second) = 1/12 ( No explanation as to why they calculated the prob. like this)
Thus answer P( white on first or white on second) = 1/13 + 1/12 = 25/156.
SO Please help with this!!
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Probability Confusion
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force5
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Well actually i feel there is something missing in the question. I dont like assuming too many things when doing a probability question atleast. The best explanation can be given by the author of the book... i am trying to understand what can the resultant be.
we are trying to find the probability of selecting a white marble on the first and second draw.
now assumptions here are we are replacing balls or not?? say we are not.
we are only selecting white marbles on both the draws. well how can that be possible when we only have one white marble in the bag.
Ok to select 1 white marble on the first draw. thats easy... 1/13
now comes the tricky part. selecting white on second draw. now its a little confusing.. here.
anyone would start doing the way you have solved. I think its perfect but the problem here is that the question doesn't clearly states the "draw"
question is to find prob of selecting the white marble on the second draw. now this statement ( to me)-- tells me that we already know that we are not selecting white on the first draw. so probability of selecting a non white ball is 1 here.
hence 1 ball gone prob of selecting a white on second draw will be 1/12 ( 12 balls left)
It is actually easy for anyone to not agree to what i have done. but a deeper insight will help you understand what i actually mean. this type of ambiguity is not generally created in GMAT and you will mostly get clear questions and thoughts.
Do feel free to ask if you are still in doubt.... but i will suggest to let it go. you have handled the question correctly.
we are trying to find the probability of selecting a white marble on the first and second draw.
now assumptions here are we are replacing balls or not?? say we are not.
we are only selecting white marbles on both the draws. well how can that be possible when we only have one white marble in the bag.
Ok to select 1 white marble on the first draw. thats easy... 1/13
now comes the tricky part. selecting white on second draw. now its a little confusing.. here.
anyone would start doing the way you have solved. I think its perfect but the problem here is that the question doesn't clearly states the "draw"
question is to find prob of selecting the white marble on the second draw. now this statement ( to me)-- tells me that we already know that we are not selecting white on the first draw. so probability of selecting a non white ball is 1 here.
hence 1 ball gone prob of selecting a white on second draw will be 1/12 ( 12 balls left)
It is actually easy for anyone to not agree to what i have done. but a deeper insight will help you understand what i actually mean. this type of ambiguity is not generally created in GMAT and you will mostly get clear questions and thoughts.
Do feel free to ask if you are still in doubt.... but i will suggest to let it go. you have handled the question correctly.
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havok
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P(1st Ball is white) = 1/13
P(2nd Ball is white) can be written as P(2nd Ball is White | 1st Ball is Blue) "Probability that 2nd ball is white given that 1st ball was Blue" - Why else would you stick your hand in the bag again?
Therefore, if you are reaching in for the 2nd ball, the first ball definitely wasn't White, and your chance is 1/12.
P(2nd Ball is white) can be written as P(2nd Ball is White | 1st Ball is Blue) "Probability that 2nd ball is white given that 1st ball was Blue" - Why else would you stick your hand in the bag again?
Therefore, if you are reaching in for the 2nd ball, the first ball definitely wasn't White, and your chance is 1/12.
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pemdas
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i think this is biased probability not taking into account information from the previous event(s).
Conditional probability would result in P=1/13 for the second draw of white, BUT biased (non-conditional probability) would result in either 0/12 or 1/12.
Conditional probability would result in P=1/13 for the second draw of white, BUT biased (non-conditional probability) would result in either 0/12 or 1/12.
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Ian Stewart
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As the question is worded, the answer is certainly 2/13. You can just imagine lining up your 13 marbles and picking the marbles at the front of the line first. There are 13 places to put the white marble, 2 of which are in the first 2 places in line, so the probability your white marble is one of the first two is 2/13 (the method used in the original post is also a perfectly good way to answer this question).vvsmart wrote:HI Experts,
Here is a question I am getting confused with (Source: 8th Grade maths):
A bag of marbles contains 12 blue marbles and 1 white marble. If you reach into the bag without looking, what is the probability of selecting the white marble on the first or second draw?
The author of that book is clearly confused, and his or her method makes no sense, as you can see by applying it to the question "If a bag contains 12 blue marbles and 1 white marble, and marbles are selected one at a time without replacement, what is the probability the white marble is one of the first 11 marbles selected?" Using the author's method, you will get an answer greater than 1, which clearly makes no sense (the answer is 11/13).
One can rephrase the question as a conditional probability question, but from the original wording, there is absolutely no reason to think conditional probability is relevant. One could ask "If a bag contains 12 blue marbles and 1 white marble, and marbles are selected one at a time without replacement, what is the probability the white marble is the second marble selected *if* the first marble is blue?" We then know something in advance about the first marble (that's our 'condition', making this a conditional probability). That does not make the math any more complicated, however, and we don't need a formula to answer - you just need to work out what the situation will be after taking account of the condition. If the first marble is blue, we have 12 marbles left, 1 of which is white, so the probability the second marble is white *if you know the first marble was blue* is 1/12.
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