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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
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Re: M0326
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19 Sep 2014, 03:40
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 6; \(C^2_3\)  # of ways to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Answer: D Kindly Correct me if i am wrong but i think that in the highlighted portion above, it should be 4 baguettes out of 7 baguettes.



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19 Sep 2014, 08:39
earnit 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 6; \(C^2_3\)  # of ways to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Answer: D Kindly Correct me if i am wrong but i think that in the highlighted portion above, it should be 4 baguettes out of 7 baguettes. Yes. Typo edited. Thank you.
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13 Jun 2015, 00:52
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 Hello. Is it possible to solve the question using the slot method? Thank you in advance.



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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!



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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
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16 Oct 2015, 10:54
I didn't understood why the baguetes were considered different.



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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?



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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 \(1/2\)



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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!
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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.
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26 May 2016, 02:02
Bunuel wrote: 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.
Hmm, I think I get the essence. If I understand it correctly: Probability of an event happening =\(\frac{"Number of ways it can happen"}{"Total number of outcomes"}\)So even though the outcome is the same (4 identical baguettes) but we can choose it in many way. Thanks for the answer Bunuel
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Hi,
Sorry for bumping in an old thread; I need to correct my understanding on concepts of probability with re: to similar questions. I ended up applying the following formula for mutually exclusive events and got it all wrong: Probability of fav. outcomes (i.e. prob. of drawing 4 Bs)= \(^6C4 * (\frac{7}{10})^4 * (\frac{3}{10})^2\)
Is it that concept of mutually exclusive events apply to scenarios with events in succession? (Ex. roll of dice, tossing of coins, weather forecast on consecutive days)
rgds.



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11 Jul 2016, 19:50
narmadadhruv wrote: Hi,
Sorry for bumping in an old thread; I need to correct my understanding on concepts of probability with re: to similar questions. I ended up applying the following formula for mutually exclusive events and got it all wrong: Probability of fav. outcomes (i.e. prob. of drawing 4 Bs)= \(^6C4 * (\frac{7}{10})^4 * (\frac{3}{10})^2\)
Is it that concept of mutually exclusive events apply to scenarios with events in succession? (Ex. roll of dice, tossing of coins, weather forecast on consecutive days)
rgds. I think the mutually exclusive events apply to concurrent events  not to a series of event that have the same characteristics. I like to think of it as when a boy proposes to a girl  to that specific girl  Yes/No is the mutually exclusive event (can't be both), however, if he proposes to another girl  this one and the previous one's answer doesn't need to be mutually exclusive (it can happen at the same time)
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12 Jul 2016, 06:25
Linhbiz wrote: narmadadhruv wrote: Hi,
Sorry for bumping in an old thread; I need to correct my understanding on concepts of probability with re: to similar questions. I ended up applying the following formula for mutually exclusive events and got it all wrong: Probability of fav. outcomes (i.e. prob. of drawing 4 Bs)= \(^6C4 * (\frac{7}{10})^4 * (\frac{3}{10})^2\)
Is it that concept of mutually exclusive events apply to scenarios with events in succession? (Ex. roll of dice, tossing of coins, weather forecast on consecutive days)
rgds. I think the mutually exclusive events apply to concurrent events  not to a series of event that have the same characteristics. I like to think of it as when a boy proposes to a girl  to that specific girl  Yes/No is the mutually exclusive event (can't be both), however, if he proposes to another girl  this one and the previous one's answer doesn't need to be mutually exclusive (it can happen at the same time) Thanks for this, u r right on, according to the definition : Mutually exclusive events cannot happen at the same time. For example: when tossing a coin, the result can either be heads or tails but cannot be both. However, it's a bit confusing for me to relate this with re: to this question. Help is much appreciated as always! Rgds. naumyuk wrote: 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 \(1/2\) Sent from my ONE A2003 using Tapatalk



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24 Aug 2016, 03:12
I think this is a highquality question and I agree with explanation.



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13 Sep 2016, 22:09
Bunuel, I din´t understand how C610 is 30 (which divided 15). Could you explain? Also, wich link you sugeest to read?



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26 Mar 2017, 10:45
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 did this way : As 4 baguettes were sold out of 7 so its 7C4/10c4=1/2 10c4=# of ways to choose 4 total out of 10 is this solution right?







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