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by amandeep.hora » Mon Feb 25, 2013 11:53 am
In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out
of the bag, what is the probability that he will have one white and one blue marble? (Assume that Bob does not replace the marbles in the bag.)


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by MoYassen23 » Mon Feb 25, 2013 12:59 pm
Hello,

There are two ways Bob can have one blue and one white marble:

1st pick - Blue --- 2nd pick - White

1st pick - White --- 2nd pick - Blue

5/10 * 2/9 = 10/90

2/10 * 5/9 = 10/90

10/90 + 10/90 = 20/90 = 2/9
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by Anurag@Gurome » Mon Feb 25, 2013 10:00 pm
amandeep.hora wrote:In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble? (Assume that Bob does not replace the marbles in the bag.)
Number of ways to select 2 marbles out of 10 marbles = 10C2 = 45

Number of ways to select one white and one blue marble = (Number of ways to select 1 white marble out of 2)*(Number of ways to select 1 blue marble out of 5) = 2*5 = 10

Hence, required probability = 10/45 = 2/9
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by Brent@GMATPrepNow » Tue Feb 26, 2013 7:46 am
amandeep.hora wrote:In a bag of marbles, there are 3 red, 2 white, and 5 blue. If Bob takes 2 marbles out
of the bag, what is the probability that he will have one white and one blue marble? (Assume that Bob does not replace the marbles in the bag.)
Whenever I'm trying to find a certain probability, it helps to stop and ask, " What exactly must occur in order for this event to happen?"

In this case, Bob can select a white marble 1st and a blue marble 2nd, or Bob can select a blue marble 1st and a white marble 2nd.

Once we have our "ors" and "ands" in order, we can create a word equation:

P(1 white and 1 blue) = P(white 1st and blue 2nd OR blue 1st and white 2nd)
= P(white 1st and blue 2nd) + P(blue 1st and white 2nd)
= [P(white 1st) x P(blue 2nd)] + [P(blue 1st) x P(white 2nd)]
= [2/10 x 5/9] + [5/10 x 2/9]
= 10/90 + 10/90
= 20/90
= [spoiler]2/9[/spoiler]

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Brent
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by ananthbi » Sun Mar 03, 2013 6:18 pm
Might be a little dumb question but why can't "1- Probability" work here? Can I not do "1 - Probability[Bob chooses red marbles]"?
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by Brent@GMATPrepNow » Sun Mar 03, 2013 6:36 pm
ananthbi wrote:Might be a little dumb question but why can't "1- Probability" work here? Can I not do "1 - Probability[Bob chooses red marbles]"?
You can use the complement here, but remember that it's:
P(Event A occurs) = 1 - P(Event A does not occur)

In this question, if event A is selecting 1 red and 1 blue, then for the event not to occur, we need either two red or two blue.

In other words, P(1 white and 1 blue) = 1 - P(NOT 1 white and 1 blue)
= 1 - P(2 red OR 2 blue)

I'll let you continue from here (it's dinner time!)

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by Anurag@Gurome » Sun Mar 03, 2013 7:03 pm
ananthbi wrote:Might be a little dumb question but why can't "1- Probability" work here? Can I not do "1 - Probability[Bob chooses red marbles]"?
Yes, you can do that.
But that trick is helpful only when the number of times "the event doesn't happen" is smaller than the number of times "the event happens".

In this case, that trick will complicate your calculation as not choosing one white and one blue doesn't mean choosing 2 red only.

P(Not 1 white and 1 blue) = P(2 red) + P (2 white) + P(2 blue) + P(1st white and 2nd red) + P(1st red and 2nd white) + P(1st blue and 2nd Red) + P(1st red and 2nd blue) ---> seven possibilities

While, P(1 white and 1 blue) = P(1st blue and 2nd white) + P(1st white and 2nd blue) ---> only two possibilities
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by ananthbi » Sun Mar 03, 2013 7:18 pm
Ahh!! That makes sense. Thanks Guys!
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