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Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. \(\frac{7}{8}\) B. \(\frac{3}{4}\) C. \(\frac{2}{3}\) D. \(\frac{5}{8}\) E. \(\frac{3}{8}\)
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16 Sep 2014, 01:29
Official Solution:Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue? A. \(\frac{7}{8}\) B. \(\frac{3}{4}\) C. \(\frac{2}{3}\) D. \(\frac{5}{8}\) E. \(\frac{3}{8}\) Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is \(\frac{5}{8}\). The same for blue marble: the probability of drawing blue marble is \(\frac{3}{8}\), the probability that for instance the 8th marble drawn is blue is still \(\frac{3}{8}\). There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results). Answer: D
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14 Nov 2014, 02:52
francisdarby wrote: Hi there, A new member here, just joined a day or two back I'm quite confused about the solution to this problem. How is it possible that the probability remains constant even when we are removing marbles without replacement? Wouldn't the denominators in the probabilities adjust for each draw? Any help would be greatly appreciated! Thanks, Francis Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? There is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it? Similar questions to practice: aboxcontains3yellowballsand5blackballsonebyone90272.htmlabagcontains3whiteballs3blackballs2redballs100023.htmleachoffourdifferentlockshasamatchingkeythekeys101553.htmlif40peoplegetthechancetopickacardfromacanister97015.htmlabagcontains3whiteballs3blackballs2redballs100023.htmlHope this helps.
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Re: M2827
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13 Jan 2018, 07:06
A simple answer using combinatorics
if the 7th slot is NOT BLUE, it is RED now ignore this slot completely. it will not affect the other arrangements. for the 7 remaining slots, we have 7 marbles = (7! arrangements) but since 4 Red repeats and 3 blue repeats, the possible arrangements = 7!/(3!4!) = 35 total arrangements without the restriction for the 7th slot = 8!(5!3!) = 56 desired probability = 35/56 = 5/8



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Re: M2827
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09 Feb 2017, 18:10
madmax1000 wrote: But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that: P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement? The reason why you got 3/8 is because you found the probability of the second draw being blue. It should be: P(b,r)+P(r,r) = 3/8*5/7 + 5/8*4/7= 5/8 Heavily recommend drawing a probability tree (try excel) to see how the probability doesn't change up until you have no more draws left. After 8 draws, the probability then drops down to zero for either color .



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17 Feb 2018, 22:58
I think this is a highquality question. I have an alternate solution, 8!/(5!*3!)  Total # of permutations  56 (i)
Of these fixing the 7th position as red  # of permutations is 7!/(4!*3!)  35(ii)
(ii)/(i) is what we want = (5/8)



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13 Nov 2014, 17:36
Hi there, A new member here, just joined a day or two back I'm quite confused about the solution to this problem. How is it possible that the probability remains constant even when we are removing marbles without replacement? Wouldn't the denominators in the probabilities adjust for each draw? Any help would be greatly appreciated! Thanks, Francis



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19 Mar 2015, 14:01
But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that: P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement?



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20 Mar 2015, 05:58
madmax1000 wrote: But if i ask the question what is P that the 2nd marble is red, wouldnt it be like that: P(b,r)+P(r,r) = 5/8*3/7 + 3/8*2/7 = 3/8
So how can P that the 7th marble is red be 5/8? How can it be for all the same if there is NO replacement? The question asks about the probability of seventh ball being not blue, while not knowing the results of previous draws. PLEASE follow the links in my previous posts to understand better.
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Re: M2827
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18 Aug 2016, 10:40
Hi Bunuel,
My understanding about this problem is that this is a case of a nonmutually exclusive event, as the execution of one event affects another, below is written what I understand from this problem.
If there are 8 marbles, and if, someone drew marbles in the manner below without replacement; B B R R R R R R
Then the respective probabilities of drawing at succession are; 3/8 2/7 5/6 4/5 3/4 2/3 1/2 1
See that above drawing leaves only a red at the seventh draw, but to reach to the seventh draw event we should also change the sample space along with the no. of favorable cases.
Do you agree?



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18 Aug 2016, 10:41
There will be more such cases to leave a nonblue at 7th draw we just need to add those cases



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18 Aug 2016, 10:52
subhajit1 wrote: Hi Bunuel,
My understanding about this problem is that this is a case of a nonmutually exclusive event, as the execution of one event affects another, below is written what I understand from this problem.
If there are 8 marbles, and if, someone drew marbles in the manner below without replacement; B B R R R R R R
Then the respective probabilities of drawing at succession are; 3/8 2/7 5/6 4/5 3/4 2/3 1/2 1
See that above drawing leaves only a red at the seventh draw, but to reach to the seventh draw event we should also change the sample space along with the no. of favorable cases.
Do you agree? Unfortunately your understanding is wrong. I suggest you to follow the links given above.
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Re: M2827
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24 Mar 2017, 11:13
Bunuel wrote: Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. \(\frac{7}{8}\) B. \(\frac{3}{4}\) C. \(\frac{2}{3}\) D. \(\frac{5}{8}\) E. \(\frac{3}{8}\)
Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is \(\frac{5}{8}\). The same for blue marble: the probability of drawing blue marble is \(\frac{3}{8}\), the probability that for instance the 8th marble drawn is blue is still \(\frac{3}{8}\). There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D Hi Bunuel the question says without replacement. It means after each draws the no of ball reduces by 1. so how the probability of drawing 7th ball remains same as probability of drawing 1st ball red
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Re: M2827
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24 Mar 2017, 11:16
daboo343 wrote: Bunuel wrote: Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. \(\frac{7}{8}\) B. \(\frac{3}{4}\) C. \(\frac{2}{3}\) D. \(\frac{5}{8}\) E. \(\frac{3}{8}\)
Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is \(\frac{5}{8}\). The same for blue marble: the probability of drawing blue marble is \(\frac{3}{8}\), the probability that for instance the 8th marble drawn is blue is still \(\frac{3}{8}\). There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D Hi Bunuel the question says without replacement. It means after each draws the no of ball reduces by 1. so how the probability of drawing 7th ball remains same as probability of drawing 1st ball red I tried to explain this here: https://gmatclub.com/forum/m28184525.html#p1442558
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31 May 2017, 09:59
This is a very good questions and hits right on the basics.



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13 Jul 2017, 10:11
For those of you who are confused by Bunuel's beautiful answer, you can arrive at the same solution using permutations.
Let's find the possible permutations for the first 6 balls, and then worry about the 7th and 8th after that
5 Reb and 1 Blue  We are arranging 6 balls of two different types. To arrange 6 objects without respect for repetition of each type, the formula is simply \(6!\). To account for repeating types, reduce \(6!\) by the factorial of the repetition number for each type.
\(\frac{6!}{5!1!}=6\)
4 Red and 2 Blue
\(\frac{6!}{4!2!}15\)
3 Red and 3 Blue
\(\frac{6!}{3!3!}=20\)
Now multiply each by the available remaining permutations of the 7th and 8th ball:
5 Reb and 1 Blue?
Only BB is left, so \(6*1=6\)
4 Reb and 2 Blue?
RB and BR are left, so \(15*2=30\)
3 Reb and 3 Blue?
Only RR is left, so \(20*1=20\)
Thus, there are \(6+30+20=56\) total arrangements
In how many of these last three options are the 7th ball red?
5 Red and 1 Blue: \(0\) 4 Red and 2 Blue: \(15\) 3 Red and 3 Blue: \(20\)
\(\frac{35}{56}=\frac{5}{8}\)
Answer D



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10 Mar 2018, 15:49
This is a simple problem disguised as something a lot more difficult and it's easy to waste a lot of time considering permutations
Consider it like this  you draw the marbles 1 at a time, therefore you have 8 positions. In all the permutations, in any one position, the probability of that marble being red is 5/8. We are considering position 7, and the probability of the marble being red is the same as all the other positions  5/8



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25 Mar 2018, 07:09
If there was a replacement of marbles, then we could say that all draws will have the same probability. Not understood why every successive draw has the same probability.



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Re: M2827
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06 Oct 2018, 04:17
Bunuel wrote: Official Solution:
Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?
A. \(\frac{7}{8}\) B. \(\frac{3}{4}\) C. \(\frac{2}{3}\) D. \(\frac{5}{8}\) E. \(\frac{3}{8}\)
Basically we need to find the probability that the seventh marble drawn is red (so not blue). Now, the initial probability of drawing red marble is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing red marble will not change for ANY successive draw: second, third, fourth, ..., seventh. Thus the probability that the seventh marble is red is \(\frac{5}{8}\). The same for blue marble: the probability of drawing blue marble is \(\frac{3}{8}\), the probability that for instance the 8th marble drawn is blue is still \(\frac{3}{8}\). There is simply no reason to believe WHY is any draw different from another (provided we don't know the other results).
Answer: D Bunuel From what part of the question stem are we judging that we are unaware of the previous outcomes of other draws? How do we differentiate this question from other replacement questions where each number of marbles would be reduced with each draw. Please explain.



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08 Oct 2018, 02:27
I used a different method using seating arrangements. Suppose there are 8 baskets placed in linear fashion, with each basket capable of holding only 1 ball. Suppose we place 8 balls 1byone in each basket Then, total number of such arrangements  8!/5!*3! = 56
Now, fix 7th basket with red ball placed in that. Now, we have to find number of arrangements of 7 balls with 4 red balls and 3 blue balls ==> 7!/4!*3! = 35
Probability = Number of desired outcomes/Total Outcomes = 35/56 or 5/8







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