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GMATPrep : On principles of probability

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GMATPrep : On principles of probability

by anirban_lax » Thu Sep 09, 2010 9:18 pm
Another one from the GMATPrep stable: Thought I got it right ...but alas!
I'm eager to know where I went wrong and would appreciate if you can please help me understand the solution.

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 paintedo n it. If one ball is to be selected at random from the box, what is the probability that the ball is either white or has an even number on it?

a) The probability that the ball will be white and will have an even number on it is 0.
b) The probability that the ball will be white minus the probability that the ball will have an even number is 0.2.

O.A. Curious to see if someone thinks the way I did. I'll share it later.

Thanks
Anirban
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by sanju09 » Fri Sep 10, 2010 2:09 am
anirban_lax wrote:Another one from the GMATPrep stable: Thought I got it right ...but alas!
I'm eager to know where I went wrong and would appreciate if you can please help me understand the solution.

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 paintedo n it. If one ball is to be selected at random from the box, what is the probability that the ball is either white or has an even number on it?

a) The probability that the ball will be white and will have an even number on it is 0.
b) The probability that the ball will be white minus the probability that the ball will have an even number is 0.2.

O.A. Curious to see if someone thinks the way I did. I'll share it later.

Thanks
Anirban

We are asked: P (W or E) = P (W) + P (E) - P (W & E) =? Where the set (W or E) is not exhaustive.

(1) P (W & E) = 0, hence, P (W or E) = P (W) + P (E) =? Insufficient

(2) P (W) - P (E) = 0.2. Insufficient

Together [spoiler]too insufficient, because we cannot answer a + b with a - b known.

IMO E[/spoiler]
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by anirban_lax » Fri Sep 10, 2010 6:27 am
Thanks Sanju! Appreciate your help!

E is indeed the OA.

I thought its C as I assumed P(WE) = P(W).P(E). Now that I rethink that has to be a wrong assumption as P(W) and P(E) are not inependent sets - a ball which is white can have even numbers written on it and thus they cannot be independent.
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by tpr-becky » Fri Sep 10, 2010 10:37 am
for or probability we need to add the probabilities of the acceptable options. here we need to add the probability of getting a white ball and the probability of getting a non-white even ball. the equation for probability is simply the number of things of the kind you want available over the total.

so here we would need (#w/total + (#red,even + #b,even)/total)

1. This is an and probability which is multiplying the probabilities so it says that (#w/total * #e/total) = 0 which means that either #w/total or #e/total is zero. So this statement tells us there are either no white or no even but we don't know which one. (BCE)

2. #w/total - #e/total =.2 here we have two unknown variable and only one equation so this one is not solveable either. (CE)

Together we only know that one of the probabilities is zero and that one of the probabilities is .2 - but we don't know which one and we still can't take the even white balls out of the equation. Therefore the answer is E.
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by beatthegmatinsept » Fri Sep 10, 2010 10:39 am
anirban_lax wrote:Thanks Sanju! Appreciate your help!

E is indeed the OA.

I thought its C as I assumed P(WE) = P(W).P(E). Now that I rethink that has to be a wrong assumption as P(W) and P(E) are not inependent sets - a ball which is white can have even numbers written on it and thus they cannot be independent.
I ended up with C too when I first attempted it few days ago. You aren't alone :)
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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