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Expert replies
by stephny » Mon Nov 11, 2013 1:13 am
hi guys,
please kindly assist.

a jar contains only red, yellow, and orange marbles. if there are 3 red, 5 yellow, and 4 orange marbles, and 3 are chosen from the jar at random without replacing any of them, what is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
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Source: — Problem Solving |

by theCodeToGMAT » Mon Nov 11, 2013 1:46 am
3R, 5Y, 4O --> 3 are chosen without replacement

Total Ways to draw 2Y, 1R = 5C2 * 3C1 = 5x4/2 * 3 = 30

Total Ways = 12C3 = 12x11x10/6 = 11*20

Probability = 30 / 11*20 = 3/22
R A H U L
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by GMATGuruNY » Mon Nov 11, 2013 4:40 am
stephny wrote:hi guys,
please kindly assist.

a jar contains only red, yellow, and orange marbles. if there are 3 red, 5 yellow, and 4 orange marbles, and 3 are chosen from the jar at random without replacing any of them, what is the probability that 2 yellow, 1 red, and no orange marbles will be chosen?
P(exactly n times) = P(one way) * total possible ways.

P(one way):
ONE WAY to select 2 yellow marbles and 1 red marble is YYR.
P(1st marble is Y) = 5/12. (Of the 12 marbles, 3 are Y.)
P(2nd marble is Y) = 4/11. (Of the 11 remaining marbles, 2 are Y.)
P(3rd marble is R) = 3/10. (Of the 10 remaining marbles, 3 are R.)
Since we want all of these events to happen, we MULTIPLY the probabilities:
5/12 * 4/11 * 3/10 = 1/22.

Total possible ways:
YYR is only ONE WAY to get exactly 2 Y's and 1 R.
Now we must account for ALL OF THE WAYS to get exactly 2 Y's and 1 R.
Any arrangement of the letters YYR represents one way to get exactly 2 Y's and 1 R..
Thus, to account for ALL OF THE WAYS to get exactly 2 Y's and 1 R, the result above must be multiplied by the number of ways to arrange the letters YYR.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical Y's:
3!/2! = 3.

Multiplying the results above, we get:
P(exactly 2 Y's and 1 R) = (1/22) * 3 = 3/22.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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by [email protected] » Tue Nov 12, 2013 12:01 am
Hi stephny,

I'm a fan of any type of Quant or Verbal shortcut that fits a given situation, BUT if you ever find yourself in a situation in which you can't "spot the shortcut", then you have to just get into the question and do the necessary work as fast as you can. This is referred to as "brute force."

In this question, we have 3 red, 5 yellow and 4 orange marbles. We're asked for the probability of grabbing 2 yellow and 1 red (in some order)?

There are 3 outcomes that satisfy this question:
YYR
YRY
RYY

Let's calculate each:
YYR = (5/12)(4/11)(3/10) = 60/1320

YRY = (5/12)(2/11)(4/10) = 60/1320

RYY = (2/12)(5/11)(4/10) = 60/1320

Total = 180/1320 = 18/132 = 6/44 = 3/22

Sometimes the "fast" way is just about mapping the possibilities and doing some math.

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