Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58434

Question Stats:
58% (02:23) correct 42% (02:43) wrong based on 197 sessions
HideShow timer Statistics
A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling) A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 58434

Official Solution:A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling) A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\) The store had 7 baguettes and 3 nonbaguettes, total of 10 loaves. 6 loaves were sold, we want to calculate the probability that exactly 4 out of theses 6 loaves were baguettes, so probability that the store sold 4 baguettes and 2 nonbaguettes, \(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\). Where: \(C^4_7\)  # of ways to choose 4 baguettes out of 7; \(C^2_3\)  # of ways to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10. Answer: D
_________________



Manager
Joined: 25 Feb 2014
Posts: 56
Concentration: Healthcare, Finance
GPA: 3.5

Re: M0326
[#permalink]
Show Tags
01 Jul 2015, 18:03
Bunuel wrote: Official Solution:
A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling)
A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\)
The store had 7 baguettes and 3 nonbaguettes, total of 10 loaves. 6 loaves were sold, we want to calculate the probability that exactly 4 out of theses 6 loaves were baguettes, so probability that the store sold 4 baguettes and 2 nonbaguettes, \(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\). Where: \(C^4_7\)  # of ways to choose 4 baguettes out of 7; \(C^2_3\)  # of ways to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Answer: D I obtained the same answer with a different numerator. I am curious if and why 6C4 is incorrect for numerator. If anyone has info, please kindly reply and thanks in advance!



Math Expert
Joined: 02 Aug 2009
Posts: 7988

Re: M0326
[#permalink]
Show Tags
01 Jul 2015, 18:51
atl12688 wrote: Bunuel wrote: Official Solution:
A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling)
A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\)
The store had 7 baguettes and 3 nonbaguettes, total of 10 loaves. 6 loaves were sold, we want to calculate the probability that exactly 4 out of theses 6 loaves were baguettes, so probability that the store sold 4 baguettes and 2 nonbaguettes, \(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\). Where: \(C^4_7\)  # of ways to choose 4 baguettes out of 7; \(C^2_3\)  # of ways to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Answer: D I obtained the same answer with a different numerator. I am curious if and why 6C4 is incorrect for numerator. If anyone has info, please kindly reply and thanks in advance! Hi, you have two points... a) why numerator cannot be taken 6C4.. 6C4 can be used in a very simple scenario where we have 6 baguettes and we have to take 4 out of them.. here it is not simple. we have to choose 4 baguettes out of 7baguettes and 2 nonbaguettes out of 3 nonbaguettes . so we have to find in how many way can we choose 4 baguettes out of 7baguettes and 2 nonbaguettes out of 3 nonbaguettes= 7C4*3C2.. b) i do not think you can get the same correct answer by taking 6C4 in numerator.. Reason; without even calculating you can see that numerator has only prime numbers upto 5 , whereas denominator has a seven , which will not cancel out and you ans will x/7y...and not 1/2.. hope it helped
_________________



Intern
Joined: 27 Dec 2011
Posts: 41
Location: Brazil
Concentration: Entrepreneurship, Strategy
GMAT 1: 620 Q48 V27 GMAT 2: 680 Q46 V38 GMAT 3: 750 Q50 V41
GPA: 3.5

Re: M0326
[#permalink]
Show Tags
16 Oct 2015, 10:54
I didn't understood why the baguetes were considered different.



Math Expert
Joined: 02 Sep 2009
Posts: 58434

Re: M0326
[#permalink]
Show Tags
17 Oct 2015, 01:00
Hugoba wrote: I didn't understood why the baguetes were considered different. Baguettes were not considered different. The solution uses \(C^x_y\) notation, which means choosing unordered group of x units out of y.
_________________



Intern
Joined: 27 Dec 2011
Posts: 41
Location: Brazil
Concentration: Entrepreneurship, Strategy
GMAT 1: 620 Q48 V27 GMAT 2: 680 Q46 V38 GMAT 3: 750 Q50 V41
GPA: 3.5

Re: M0326
[#permalink]
Show Tags
19 Oct 2015, 05:42
Hi Bunuel,
Thanks for the reply. It may seem a stupid, but I'm confused by the way the problem was worded Consider the following set A:
A= {b,b,b,b,b,b,b}
Now consider the following subsets C and D of A:
C={b,b,b,b}; D={b,b,b,b}
C was obtained from A by selecting the underlined elements of A as follows:
A= {b,b,b,b,b,b,b}
D was obtained from A by selecting the underlined elements of A as follows:
D= {b,b,b,b,b,b,b}
Aren't sets C and D (by the purpose of the problem) the same?



Current Student
Joined: 29 Apr 2015
Posts: 26
Location: Russian Federation
GPA: 4

Bunuel wrote: A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling)
A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\) I have different approach  to use probability theory. Let's get a product of all possible descreate probabilities. What is probability to get 1 baguette? It's 7/10. If we get 1 baguette and want to get one more baguette, then probability to get two baguettes is product of 7/10 * 6/9. Move on and on, eventually we will get the following: \(7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5 = 1/30\) It's a probability to get 4 baguettes and 2 loaves of other kind. But probability that we got works only if we get loaves in exactly following order: baguette + baguette + baguette + baguette + nonbaguette + nonbaguette But there can be another order. In order to count a number of all possible orders we have to find \(6!/4!*2! = 15\) So, we have to summ gotten probability 15 times. It means \(1/30 * 15 = 1/2\) So the sought probability is \(1/2\)



Intern
Joined: 08 Jul 2015
Posts: 40
GPA: 3.8
WE: Project Management (Energy and Utilities)

Hugoba wrote: Hi Bunuel,
Thanks for the reply. It may seem a stupid, but I'm confused by the way the problem was worded Consider the following set A:
A= {b,b,b,b,b,b,b}
Now consider the following subsets C and D of A:
C={b,b,b,b}; D={b,b,b,b}
C was obtained from A by selecting the underlined elements of A as follows:
A= {b,b,b,b,b,b,b}
D was obtained from A by selecting the underlined elements of A as follows:
D= {b,b,b,b,b,b,b}
Aren't sets C and D (by the purpose of the problem) the same? I have the same concern as Hugoba  after research out, the OA may not be correct if all baguettes are considered the same. ( Bunuel, kindly correct if my following explanation flawed) First, back to basic, the definition of combination: "A kcombination of a set S is a subset of k distinct elements of S. If the set has n elements, the number of kcombinations is equal to the binomial coefficient or \(C\binom{n}{k}\)"The key thing to note here is the distinct elements  that means \(C\binom{n}{k}\) formula can only apply to the set if its elements are different, if there are some repetition of elements (or identical elements) in the group, \(C\binom{n}{k}\) formula cannot be applied.Example: how many different combination of choossing 4 baguettes out of 7 identical baguettes (without considering the order)? Ans: \(C\binom{7}{4}\)?  Nope! No matter what way you choose, from left to right, center to left to right, first of last, etc etc  you'll always end with the identical 4 baguettes > that means there is only 1 expected result from this > so the number of different combination is always 1 Back to the problem, how I tackle this?1. You buy 6 out of 10 (7B+3S) (Sfor something else)  all baguettes are the same, and all something else are the same. 2. How many possible way you can buy 6 loads?  only 4 0S + 6B 1S + 5B 2S + 4B > this is a desirable outcome3S + 3B So the probability of desirable outcome is \(p = 1/4 = 0.25\)That's my understanding, hope to hear more from you guys!
_________________
[4.33] In the end, what would you gain from everlasting remembrance? Absolutely nothing. So what is left worth living for? This alone: justice in thought, goodness in action, speech that cannot deceive, and a disposition glad of whatever comes, welcoming it as necessary, as familiar, as flowing from the same source and fountain as yourself. (Marcus Aurelius)



Math Expert
Joined: 02 Sep 2009
Posts: 58434

Re: M0326
[#permalink]
Show Tags
26 May 2016, 01:17
Linhbiz wrote: Hugoba wrote: Hi Bunuel,
Thanks for the reply. It may seem a stupid, but I'm confused by the way the problem was worded Consider the following set A:
A= {b,b,b,b,b,b,b}
Now consider the following subsets C and D of A:
C={b,b,b,b}; D={b,b,b,b}
C was obtained from A by selecting the underlined elements of A as follows:
A= {b,b,b,b,b,b,b}
D was obtained from A by selecting the underlined elements of A as follows:
D= {b,b,b,b,b,b,b}
Aren't sets C and D (by the purpose of the problem) the same? I have the same concern as Hugoba  after research out, the OA may not be correct if all baguettes are considered the same. ( Bunuel, kindly correct if my following explanation flawed) First, back to basic, the definition of combination: "A kcombination of a set S is a subset of k distinct elements of S. If the set has n elements, the number of kcombinations is equal to the binomial coefficient or \(C\binom{n}{k}\)"The key thing to note here is the distinct elements  that means \(C\binom{n}{k}\) formula can only apply to the set if its elements are different, if there are some repetition of elements (or identical elements) in the group, \(C\binom{n}{k}\) formula cannot be applied.Example: how many different combination of choossing 4 baguettes out of 7 identical baguettes (without considering the order)? Ans: \(C\binom{7}{4}\)?  Nope! No matter what way you choose, from left to right, center to left to right, first of last, etc etc  you'll always end with the identical 4 baguettes > that means there is only 1 expected result from this > so the number of different combination is always 1 Back to the problem, how I tackle this?1. You buy 6 out of 10 (7B+3S) (Sfor something else)  all baguettes are the same, and all something else are the same. 2. How many possible way you can buy 6 loads?  only 4 0S + 6B 1S + 5B 2S + 4B > this is a desirable outcome3S + 3B So the probability of desirable outcome is \(p = 1/4 = 0.25\)That's my understanding, hope to hear more from you guys! Yes, all baguettes and nonbaguettes are identical but the number of ways to choose 4 baguettes out of 6 is not 1. Consider 7 baguettes in a row, we want to choose 4 out of them: we can choose 1st, 2nd 3rd and 4th, OR 1st, 2nd 3rd and 5th, OR 1st, 2nd 3rd and 5th, ... That's how we are getting 35 combinations. H Hope it's clear.
_________________



Senior Manager
Joined: 31 Mar 2016
Posts: 375
Location: India
Concentration: Operations, Finance
GPA: 3.8
WE: Operations (Commercial Banking)

Re M0326
[#permalink]
Show Tags
24 Aug 2016, 03:12
I think this is a highquality question and I agree with explanation.



Manager
Joined: 22 Jun 2017
Posts: 168
Location: Argentina

Re M0326
[#permalink]
Show Tags
27 Feb 2019, 09:19
I think this is a highquality question and I agree with explanation. I solve this problem in another way. 6C4 * 7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/8 =1/2
_________________
The HARDER you work, the LUCKIER you get.



Intern
Joined: 18 Jan 2018
Posts: 38
Location: India
Concentration: Finance, Marketing
GPA: 3.98

Re: M0326
[#permalink]
Show Tags
12 Mar 2019, 23:10
Hi BunuelCan't we solve this question using the binomial situation (i.e by applying the formula 6C4 * (7/10)^4 * 3/10^2)?



Manager
Joined: 19 Jan 2018
Posts: 67

Re: M0326
[#permalink]
Show Tags
11 Aug 2019, 16:38
Bunuel wrote: A store had 10 loaves of bread, out of which 7 were baguettes. If the store sold 6 loaves of bread, what is the probability that the store sold exactly 4 baguettes out of these 6 loaves? (Assume that every loaf has an equal chance of selling)
A. \(\frac{2}{5}\) B. \(\frac{3}{5}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) E. \(\frac{4}{7}\) We know that there's 10 loaves of bread, 7 of which are baguettes, which means that 3 of them are not baguettes. We can set up a simple probability for each loaf (Blue is the probability that exactly 4 baguettes are selected, while the red is the probability that the other type of bread is selected) \(\frac{7}{10}\) * \(\frac{6}{9}\) * \(\frac{5}{8}\) * \(\frac{4}{7}\) * \(\frac{3}{6}\) * \(\frac{2}{5}\)After you simplify that, you get \(\frac{1}{30}\), but you have to use \(C(6, 4)\), because you can select the 4 Baguettes in different orders c(6, 4) = 15 and you take \(\frac{1}{30}\) * \(15\) = \(\frac{1}{2}\) Answer D is correct










