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by tmontgomery » Fri Sep 04, 2015 10:13 am
A bag contains 6 red marbles and 4 blue marbles. If a marble is drawn with replacement, what is the minimum number of draws required so that there is at least a 70% probability of drawing a blue marble.
A. 6
B. 4
C. 3
D. 2
E. 5
Last edited by tmontgomery on Fri Sep 04, 2015 1:35 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Fri Sep 04, 2015 10:51 am
Please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions.

Also, be sure to include all 5 answer choices. In many cases, the fastest approach involves working with the answer choices.

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by Matt@VeritasPrep » Fri Sep 04, 2015 5:10 pm
To find the probability of drawing a blue marble, we can simply compute 1 - (Probability of drawing all red marbles).

If we only draw one marble, this gives us 1 - (6/10), or 3/5. Not high enough.

If we draw two marbles, this gives us 1 - (6/10)*(5/9), or 2/3. Still not high enough!

If we draw three marbles, this gives us 1 - (6/10)*(5/9)*(4/8), or 5/6. Success!
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by sandipgumtya » Sun Sep 06, 2015 4:17 am
Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?
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by [email protected] » Sun Sep 06, 2015 1:02 pm
Hi sandipgumtya,

When it comes to probability questions, there are only two things that you can calculate: what you WANT to have happen and what you DON'T want to have happen. Since the probability of what you WANT + the probability of what you DON'T WANT will always add up to the number 1, in certain questions, it's actually easier to calculate what you DON'T WANT, then subtract that fraction from the number 1 (to figure out the probability of what you DO WANT occurring).

We can use this approach on this prompt...

IF...you draw just one marble, the probability of it being red is 6/10, so the probability of it being blue is 1 - 6/20 = 4/10 = 40%; which is not high enough.

IF...you draw two marbles, the probability of them both being red is (6/10)(5/9) = 30/90 = 1/3, so the probability of at least one being blue is 1 - 1/3 = 2/3 = 66 2/3%; which is not high enough.

IF...you draw three marbles, the probability of all three being red is (6/10)(5/9)(4/8) = 1/6, so the probability of at least one being blue is 1 - 1/6 = 5/6 = 83 1/3%; which IS high enough.

Final Answer: C

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by Matt@VeritasPrep » Tue Sep 08, 2015 2:44 am
sandipgumtya wrote:Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?
Sure!

When I'm picking these marbles, there are (on some level) only two possible results.

Result #1:: I get all red marbles
Result #2:: I DON'T get all red marbles

Since these are the only two possibilities, we must have

(Result #1) + (Result #2) = 100%

This implies that

Result #2 = 100% - (Result #1), or

Result #2 = 1 - Result #1

From here, we see that

(Getting at least one blue marble) = 1 - (Getting all red marbles)
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by Brent@GMATPrepNow » Tue Sep 08, 2015 5:34 am
tmontgomery wrote:A bag contains 6 red marbles and 4 blue marbles. If a marble is drawn with replacement, what is the minimum number of draws required so that there is at least a 70% probability of drawing a blue marble.
A. 6
B. 4
C. 3
D. 2
E. 5
We can reword the question as "What is the minimum number of draws required so that the probability of drawing ALL RED marbles is LESS THAN 30%?"

On ANY draw, P(red marble) = 6/10

Now try different values.

2 draws
So, P(red marbles on all 2 draws) = (6/10)(6/10) = 36/100 = 36%
This probability is GREATER than 30% so keep going.

3 draws
So, P(red marbles on all 3 draws) = (6/10)(6/10)(6/10) = 216/1000 = 21.6%
This probability is LESS than 30%. PERFECT!

Answer: C

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by gmatdriller » Tue Sep 08, 2015 10:49 pm
[email protected] wrote:Hi sandipgumtya,

When it comes to probability questions, there are only two things that you can calculate: what you WANT to have happen and what you DON'T want to have happen. Since the probability of what you WANT + the probability of what you DON'T WANT will always add up to the number 1, in certain questions, it's actually easier to calculate what you DON'T WANT, then subtract that fraction from the number 1 (to figure out the probability of what you DO WANT occurring).

We can use this approach on this prompt...

IF...you draw just one marble, the probability of it being red is 6/10, so the probability of it being blue is 1 - 6/20 = 4/10 = 40%; which is not high enough.

IF...you draw two marbles, the probability of them both being red is (6/10)(5/9) = 30/90 = 1/3, so the probability of at least one being blue is 1 - 1/3 = 2/3 = 66 2/3%; which is not high enough.

IF...you draw three marbles, the probability of all three being red is (6/10)(5/9)(4/8[/u]) = 1/6, so the probability of at least one being blue is 1 - 1/6 = 5/6 = 83 1/3%; which IS high enough.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich


Hi Rich,

The question says "with replacements"..what do you about the part in BLUE, BOLD?
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by Max@Math Revolution » Wed Sep 09, 2015 7:09 am
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


A bag contains 6 red marbles and 4 blue marbles. If a marble is drawn with replacement, what is the minimum number of draws required so that there is at least a 70% probability of drawing a blue marble.
A. 6
B. 4
C. 3
D. 2
E. 5

The probability that at least one marble is blue when drawn out one at a time = 1 - probability that all the marbles are red.

probability that all the marbles are red.

once : 6C1/10C1=4/6=66.7% -->probability that at least one marble is blue when drawn out one at a time= 1-66.7%=33.3%<70%
twice:(6C1/10C1)(6C1/10C1)=9/25=36% -->probability that at least one marble is blue when drawn out one at a time= 1-36%=64%<70%
third times:(6C1/10C1)(6C1/10C1)(6C1/10C1)=27/125=21.6% --> probability that at least one marble is blue when drawn out one at a time= 1-21.6%=78.4%>70%. Thus the answer is 3, C.


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by sandipgumtya » Wed Sep 09, 2015 9:26 am
thanks Matt.
@Brent:The question says "with replacement"but i think u considered without.can u plz check and confirm.
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by Brent@GMATPrepNow » Wed Sep 09, 2015 9:32 am
sandipgumtya wrote:thanks Matt.
@Brent:The question says "with replacement"but i think u considered without.can u plz check and confirm.
If we replace each marble after we select it, then P(red marble) = 6/10 every time.
With replacement, we ALWAYS have 10 marbles to choose from and they are ALWAYS 6 red among them.

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by [email protected] » Wed Sep 09, 2015 9:58 am
Hi gmatdriller,

Thanks for pointing out the error; conceptually though, the math is still the same...

IF...you draw just one marble, the probability of it being red is 6/10, so the probability of it being blue is 1 - 6/10 = 4/10 = 40%; which is not high enough.

IF...you draw two marbles, the probability of them both being red is (6/10)(6/10) = 36/100, so the probability of at least one being blue is 1 - 36/100 = 64/100 = 64%; which is not high enough.

IF...you draw three marbles, the probability of all three being red is (6/10)(6/10)(6/10)= 216/1000, so the probability of at least one being blue is 1 - 216/1000 = 784/1000 = 78.4%; which IS high enough.

Final Answer: C

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by Amrabdelnaby » Tue Nov 17, 2015 5:28 pm
There's a very easy way to do it.

Consider the following: since the question asks us about the least number of withdraws and replacements we can make in order to achieve 7 marbles we will assume that every time we withdraw a marble it will be red,

now we have the following ratio of red to blue marbles

Red Blue Total
Before making any withdraws 6 4 10
After withdrawing one red marble and switching it with blue 5 5 10
After withdrawing 2 red marbles and changing them to blue 4 6 10
After withdrawing 3 red marbles and changing them to blue 3 7 10

BINGO!
sandipgumtya wrote:Hi Matt,
Didn't get how r u getting the probability as 1-(getting all red marble)??Can u pl explain once more?
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