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A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbles, and 5 magenta marbles. First, four marbles are removed at random, without replacement. Next, three more marbles are removed at random, without replacement. What is the probability that the remaining marbles are the same color?

Total outcomes 10c4 X 6c3. This equals total of 4200 outcomes. Then we will take out 4 out of 7 in 7c4 ways an then 3 out of 3 in 1 ways. Thus probability equals (7c4X 3c3) / 4200. Thus 1/120. Therefor C is the answer.

Basically we are picking 7 marbles out of 10 and the number ways to do that is 10C7 = 120 Now basically for having same color of marbles we have 2 options, either all cerulean marbles or 3 of the magenta are remaining.

so 2 alabaster + 5 magenta or 3 cerulean + 2 alabaster + 2 magenta

Total outcomes 10c4 X 6c3. This equals total of 4200 outcomes. Then we will take out 4 out of 7 in 7c4 ways an then 3 out of 3 in 1 ways. Thus probability equals (7c4X 3c3) / 4200. Thus 1/120. Therefor C is the answer.

Basically we are picking 7 marbles out of 10 and the number ways to do that is 10C7 = 120 Now basically for having same color of marbles we have 2 options, either all cerulean marbles or 3 of the magenta are remaining.

so 2 alabaster + 5 magenta or 3 cerulean + 2 alabaster + 2 magenta

Basically we are picking 7 marbles out of 10 and the number ways to do that is 10C7 = 120 Now basically for having same color of marbles we have 2 options, either all cerulean marbles or 3 of the magenta are remaining.

so 2 alabaster + 5 magenta or 3 cerulean + 2 alabaster + 2 magenta

Dear, How is 2a*5m =1*1? 3c*2a = 1*1? I really like your approach.. but i don't understand this part.. please explain.. thank you so much!

Number of ways of selecting 2 alabaster marbles and 5 magenta marbles = 1 Number of ways of selecting 3 cerulean , 2alabaster and 2 out of 5 magenta = 1*1*5C2 = 10

A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbles, and 5 magenta marbles. First, four marbles are removed at random, without replacement. Next, three more marbles are removed at random, without replacement. What is the probability that the remaining marbles are the same color?

(A) 1/7 (B) 1/8 (C) 1/120 (D) 7/120 (E) 11/120

Total marbles = 10

After removal of 4 and then 3 marbles , remaining marbles = 10-4-3 = 3

The only way 3 marbles can be of the same color is when the marbles are either cerulean or magenta.

This question can be reinterpreted as : what is the probability of drawing 3 cerulean or 3 magenta marbles from 2 alabaster marbles, 3 cerulean marbles, and 5 magenta marbles.

No of ways of drawing 3 marbles out of 10 = 10C3 = 120

No of ways of drawing 3 Cerulean marbles out of 3 = 3C3 =1

No of ways of drawing 3 Magenta marbles out of 5 = 5C3 =10

Total outcomes 10c4 X 6c3. This equals total of 4200 outcomes. Then we will take out 4 out of 7 in 7c4 ways an then 3 out of 3 in 1 ways. Thus probability equals (7c4X 3c3) / 4200. Thus 1/120. Therefor C is the answer.

The sequential removal of marbles is a total red herring and you can ignore it. (This happens sometimes in combinatorics problems! To spot cases like this, think about whether what the problem is telling you is actually different from the simpler case. Is removing three marbles then four marbles (without looking at them) different from removing 7 marbles? Nope - that's like saying that eating four cookies then three cookies right afterwards, is somehow different from eating seven cookies all in a row.)

Now let's build up the probability.

probability = good cases / total cases

A 'good' case is one where the three marbles remaining are all the same color. They can't all be alabaster, since there definitely aren't enough alabaster marbles. They could all be cerulean; since there are exactly three cerulean marbles, there's only one way for that to happen. They could also all be magenta. Suppose that the magenta marbles were labeled A, B, C, D, and E. How many possible cases are there where three magenta marbles are left over?

ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE

That's ten possibilities. So, there are eleven possible cases where the three marbles left over are all the same color.

I stopped at this point and picked the one answer choice that has 11 in the numerator.

That's what you should do on the GMAT! But if you're just curious about the math, here's where the 120 in the denominator comes from. How many total cases are there? That is, how many ways can 3 marbles be left over, whether they're all the same or not? That's '10 choose 3', or 10!/(7!*3!), which equals 120.
_________________

Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

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