A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbl

Question

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

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

quantumliner · Accepted Answer

The last three marbles can be either cerulean marbles or magenta marbles.

So how if we reverse the selection, i.e First three marbles are selected without replacement, such that the first three marbles are same.

Case 1 : First three marbles are Cerulean = \frac{3}{10} * \frac{2}{9} * \frac{1}{8} = \frac{6}{720}

Case 2:  First three marbles are Magenta =  \frac{5}{10} * \frac{4}{9} * \frac{3}{8} = \frac{60}{720}

Total Probability =  \frac{6}{720} + \frac{60}{720} = \frac{66}{720} = \frac{11}{120}

Answer is E

kumarparitosh123 · Answer

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.

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ganj28 · 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

2a* 5m is 1 combination = 1* 1 = 1
3c * 2a * 5C2 = 1 * 1* 10

So totally 11 out 120 = 11/120

(E)

pclawong · Answer

Dear,
Would you explain what is 7c4 and 3c3? 
Thank you so much.

pclawong · Answer

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!

ganj28 · Answer

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

Hope its clear.

darn · Answer

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

(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

Therefore probability = (10+1) / 120 = 11/120

kumarparitosh123 · Answer

No my approach and solution for this problem are flawed.

Thanks for pointing it.

Regards

Sent from my Lenovo TAB S8-50LC using  GMAT Club Forum mobile app

ccooley · 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.

ScottTargetTestPrep · Answer

Solution:
 
We see that after 7 of the 10 marbles are removed, if the remaining 3 marbles are the same color, then they must be either all cerulean or all magenta, as these are the only two colors that have at least 3 marbles.

Case 1:  The last 3 marbles are cerulean

This means the first 7 marbles that are removed consist of 2 alabaster marbles and 5 magenta marbles. The probability that these marbles are removed is:

(2C2 x 5C5) / (10C7) = (1 x 1) / 120 = 1/120

Case 2:  The last 3 marbles are magenta

This means the first 7 marbles that are removed consist of 2 alabaster marbles, 3 cerulean marbles and 2 magenta marbles. The probability that these marbles are removed is:

(2C2 x 3C3 x 5C2) / (10C7) = (1 x 1 x 10) / 120 = 10/120

Therefore, the total probability is 1/120 + 10/120 = 11/120.

Answer: E

bumpbot · Answer

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A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbl

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A magical leather pouch contains 2 alabaster marbles, 3 cerulean marbl

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