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A store buys 10 loaves of bread, including 7 baguettes. If that day, it sells 6 loaves of bread one by one, what is the probability of the store selling exactly 4 baguettes among the 6 loaves sold? (Assume that every loaf has an equal chance of selling) A) 2/5 B) 3/5 C) 2/3 D) 1/2 E) 4/7 In the solution am unable to understand why are we selecting 2 out of 3 loaves as a part of our favorable outcome, the other part with 7c4 (i.e. for baguettes) is understandable. Can somebody help explain this.
Thanks!



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Re: M03 #26 [#permalink]
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21 Jun 2010, 00:19
we have 10 loaves:7 baguettes and 3 "non baguettes" we select 4 baguettes out of 7 baguettes and we select 2 other breads from "3 non baguettes" . 4+2 makes 6 loaves



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Re: M03 #26 [#permalink]
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21 Jun 2010, 00:55
Hey Can anyone help me for the solution of this problem........



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Re: M03 #26 [#permalink]
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21 Jun 2010, 01:44
abhi758 wrote: A store buys 10 loaves of bread, including 7 baguettes. If that day, it sells 6 loaves of bread one by one, what is the probability of the store selling exactly 4 baguettes among the 6 loaves sold? (Assume that every loaf has an equal chance of selling)
A) 2/5 B) 3/5 C) 2/3 D) 1/2 E) 4/7
In the solution am unable to understand why are we selecting 2 out of 3 loaves as a part of our favorable outcome, the other part with 7c4 (i.e. for baguettes) is understandable. Can somebody help explain this.
Thanks! You need to determine the probability of selling exactly 4 baguettes amongst 6 sold. We are interested in determining the following ratio: Total Outcomes = Favorable Outcomes/Total Outcomes 1. Total Outcomes: Determine the total number of ways in which 6 loaves of bread can be sold out of 10. 10 breads are available for sale, of which 6 are sold. So there are 10C6 ways in which 6 breads can be sold out of 10. 2. Favorable Outcomes: Determine the total number of ways in which 4 loaves of baguettes can be sold amongst 6 sold. There are 7 loaves of baguette's available, of which 4 are sold. So there are 7C4 ways of picking 4 baguette's amongst 7. There are 3 other types of breads available, of which 2 are sold. So there are 3C2 ways of selling 2 breads that are not baguettes. The total favorable outcomes, selling 4 out of 6 breads, is 7C4*3C2 3. Probability: Hence the probability of selling exactly 4 baguettes out of 6 total sold is (7C4 *3C2)/10C6



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Re: M03 #26 [#permalink]
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21 Jun 2010, 02:39
abhi758 wrote: A store buys 10 loaves of bread, including 7 baguettes. If that day, it sells 6 loaves of bread one by one, what is the probability of the store selling exactly 4 baguettes among the 6 loaves sold? (Assume that every loaf has an equal chance of selling)
A) 2/5 B) 3/5 C) 2/3 D) 1/2 E) 4/7
In the solution am unable to understand why are we selecting 2 out of 3 loaves as a part of our favorable outcome, the other part with 7c4 (i.e. for baguettes) is understandable. Can somebody help explain this.
Thanks! There are 7 baguettes and 3 nonbaguettes, total 10 loaves. 6 loaves were sold, we want to calculate the probability that exactly 4 among theses 6 loaves were baguettes, so probaility 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 way to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10. Hope it's clear.
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Re: M03 #26 [#permalink]
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21 Jun 2010, 08:21
Thanks Bunnel, Praetorian & Pipp! Nice explanations..



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Re: M03 #26 [#permalink]
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30 Jun 2011, 13:31
Bunuel wrote: \(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\).
Where: \(C^4_7\)  # of ways to choose 4 baguettes out of 6; \(C^2_3\)  # of way to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Hope it's clear.
Hi Bunuel; need help! doesn't the statement in the questions 'one by one' have any impact on way we work out the solution based on reducing sequential decision. something like following 3/10 x 2/9 x 7/8 x 6/7 x 5/6 x 4/5 = 1/30 first two is successive probability of nonbaguette and then 4 successive baguette. but yes, at the same time this need not be the order and B or NB can get sold of on any order. so we need Combination formula to work it out (not permutation). . . so i don't know if i am thinking right?? essentially what is the impact of 'one by one' Tx Target760
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Re: GC m03q26 Loaves of bread [#permalink]
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04 Feb 2012, 03:56
I am doing something here.... but I do not know where:
10c6=210 7c4=35 3c2=3 My final result is = 3/14..... I am sure I am doing a calculus mistake.... can anyone point it out?
Thanks



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Re: GC m03q26 Loaves of bread [#permalink]
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04 Feb 2012, 05:43



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Re: M03 Q26 [#permalink]
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10 May 2012, 12:20
I was thinking along these lines and i found that this question could be wrong. Please go through my approach and let me why this could be wrong:
We have 10 loaves of bread of which 7 are baguettes and 3 others(could be all three different kinds,two of same kind or all of the same kind).
This first approach assumes that the 3 remaining loaves are all of the same kind.
Since we need the probability of exactly 4 baguettes(which are indistinguishable from the perspective of the sale, do we care which baguette got sold i.e the short one the long one the burnt one??) and two from the rest of the 3 loaves which are indistinguishable since they are all of the same type of bread.
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes.( since all indistinguishable there is only one way of selecting 3 others and 3 baguettes.) 2) 2 Loaves of Other bread and 4 baguettes.(Only one way of this happening same logic as stated above) 3) 1 Loaf of Other bread and 5 baguettes.(Only one way of this happening same logic as stated above) 4) 6 baguettes.(Only one way of this happening same logic as stated above)
So the probability is 1 favorable outcome by 4 total outcomes = 1/4.
If assume that the other 3 loaves of bread are all of different types, the answer would be :
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(3 ways of this happening since there are 3 ways of selecting 2 different loaves of bread from 3) 3) 1 Loaf of Other bread and 5 baguettes.(3 ways of this happening since there are 3 ways of selecting 1 different loaf of bread from 3) 4) 6 baguettes.(Only one way of this happening)
So the probability is 3 favorable outcomes by 8 total outcomes = 3/8.
Similarly we could calculate for the scenario when two of three other loaves of bread are of the same kind (indistinguishable). Lets call the other three loaves as O1,O2,O2. So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(2 ways of this happening since O1,O2 or O2,O2 assuming the other) 3) 1 Loaf of Other bread and 5 baguettes.(2 ways of this happening O1 or O2) 4) 6 baguettes.(Only one way of this happening)
So the probability is 2 favorable outcomes by 6 total outcomes = 1/3.
So the required piece of information is whether the 3 loaves of bread other than the baguettes are indistinguishable or not.
My entire solution relies on this below point that:
We are selecting a subset of indistinguishable objects from a set indistinguishable objects. There is only one way of doing this. The order in which the objects are selected does not matter since they are indistinguishable objects. The problem is like selecting 6 letters(and the order of this 6 letter is not relevant) from a set of letters which is {B,B,B,B,B,B,B,O,O,O}.
BBBBOO is the same as OBBOBB since at the end of the day 4 baguettes and 2 others got sold.
I hope i'm making sense. Please let me know why this approach is wrong. What assumption on my part is incorrect?



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Re: M03 Q26 [#permalink]
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17 Jun 2012, 02:34
Hi ,
Can someone please recommend how to solve this problem using conditional probability method?



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Re: M03 #26 [#permalink]
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17 Jun 2012, 03:00
Target760 wrote: Bunuel wrote: \(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\).
Where: \(C^4_7\)  # of ways to choose 4 baguettes out of 7; \(C^2_3\)  # of way to choose 2 nonbaguettes out of 3; \(C^6_{10}\)  total # of ways to choose 6 loaves out of 10.
Hope it's clear.
Hi Bunuel; need help! doesn't the statement in the questions 'one by one' have any impact on way we work out the solution based on reducing sequential decision. something like following 3/10 x 2/9 x 7/8 x 6/7 x 5/6 x 4/5 = 1/30 first two is successive probability of nonbaguette and then 4 successive baguette. but yes, at the same time this need not be the order and B or NB can get sold of on any order. so we need Combination formula to work it out (not permutation). . . so i don't know if i am thinking right?? essentially what is the impact of 'one by one' Tx Target760 Hi, Yes, you can do this problem by using one by one approach also. But that wont be necessary. Probability of picking one by one in any given order = 1/30 (you got this) Total number of orders in which one can pick = 6!/4!2! = 15 (Number of ways of arranging 4B and 2NB) Required probability = 15 * 1/30 = 1/2 Hope it helps.



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Re: M03 Q26 [#permalink]
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17 Jun 2012, 03:25
go0ose wrote: I was thinking along these lines and i found that this question could be wrong. Please go through my approach and let me why this could be wrong:
We have 10 loaves of bread of which 7 are baguettes and 3 others(could be all three different kinds,two of same kind or all of the same kind).
This first approach assumes that the 3 remaining loaves are all of the same kind.
Since we need the probability of exactly 4 baguettes(which are indistinguishable from the perspective of the sale, do we care which baguette got sold i.e the short one the long one the burnt one??) and two from the rest of the 3 loaves which are indistinguishable since they are all of the same type of bread.
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes.( since all indistinguishable there is only one way of selecting 3 others and 3 baguettes.) 2) 2 Loaves of Other bread and 4 baguettes.(Only one way of this happening same logic as stated above) 3) 1 Loaf of Other bread and 5 baguettes.(Only one way of this happening same logic as stated above) 4) 6 baguettes.(Only one way of this happening same logic as stated above)
So the probability is 1 favorable outcome by 4 total outcomes = 1/4.
If assume that the other 3 loaves of bread are all of different types, the answer would be :
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(3 ways of this happening since there are 3 ways of selecting 2 different loaves of bread from 3) 3) 1 Loaf of Other bread and 5 baguettes.(3 ways of this happening since there are 3 ways of selecting 1 different loaf of bread from 3) 4) 6 baguettes.(Only one way of this happening)
So the probability is 3 favorable outcomes by 8 total outcomes = 3/8.
Similarly we could calculate for the scenario when two of three other loaves of bread are of the same kind (indistinguishable). Lets call the other three loaves as O1,O2,O2. So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(2 ways of this happening since O1,O2 or O2,O2 assuming the other) 3) 1 Loaf of Other bread and 5 baguettes.(2 ways of this happening O1 or O2) 4) 6 baguettes.(Only one way of this happening)
So the probability is 2 favorable outcomes by 6 total outcomes = 1/3.
So the required piece of information is whether the 3 loaves of bread other than the baguettes are indistinguishable or not.
My entire solution relies on this below point that:
We are selecting a subset of indistinguishable objects from a set indistinguishable objects. There is only one way of doing this. The order in which the objects are selected does not matter since they are indistinguishable objects. The problem is like selecting 6 letters(and the order of this 6 letter is not relevant) from a set of letters which is {B,B,B,B,B,B,B,O,O,O}.
BBBBOO is the same as OBBOBB since at the end of the day 4 baguettes and 2 others got sold.
I hope i'm making sense. Please let me know why this approach is wrong. What assumption on my part is incorrect? Hi, I should say, you made a few funny assumptions. "We are selecting a subset of indistinguishable objects from a set indistinguishable objects. There is only one way of doing this. The order in which the objects are selected does not matter since they are indistinguishable objects." Imagine a box containing 1000000 marbles. There are 999998 identical red marbles and 2 identical white marbles. What is the probability of picking 2 red marbles and 2 white marbles if you pick 4 marbles from the box?? According to your logic, it would be 1... Got it??



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Re: M03 #26 [#permalink]
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20 Jul 2012, 21:36
I understand all of the math here except why the number of ways to choose 4 baguettes out of 6 is C^4_7. Shouldn't it be C^4_6?



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Re: M03 #26 [#permalink]
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21 Jul 2012, 01:44



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Re: M03 Q26 [#permalink]
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19 May 2014, 22:21
go0ose wrote: I was thinking along these lines and i found that this question could be wrong. Please go through my approach and let me why this could be wrong:
We have 10 loaves of bread of which 7 are baguettes and 3 others(could be all three different kinds,two of same kind or all of the same kind).
This first approach assumes that the 3 remaining loaves are all of the same kind.
Since we need the probability of exactly 4 baguettes(which are indistinguishable from the perspective of the sale, do we care which baguette got sold i.e the short one the long one the burnt one??) and two from the rest of the 3 loaves which are indistinguishable since they are all of the same type of bread.
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes.( since all indistinguishable there is only one way of selecting 3 others and 3 baguettes.) 2) 2 Loaves of Other bread and 4 baguettes.(Only one way of this happening same logic as stated above) 3) 1 Loaf of Other bread and 5 baguettes.(Only one way of this happening same logic as stated above) 4) 6 baguettes.(Only one way of this happening same logic as stated above)
So the probability is 1 favorable outcome by 4 total outcomes = 1/4.
If assume that the other 3 loaves of bread are all of different types, the answer would be :
So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(3 ways of this happening since there are 3 ways of selecting 2 different loaves of bread from 3) 3) 1 Loaf of Other bread and 5 baguettes.(3 ways of this happening since there are 3 ways of selecting 1 different loaf of bread from 3) 4) 6 baguettes.(Only one way of this happening)
So the probability is 3 favorable outcomes by 8 total outcomes = 3/8.
Similarly we could calculate for the scenario when two of three other loaves of bread are of the same kind (indistinguishable). Lets call the other three loaves as O1,O2,O2. So the possible outcomes of 6 loaves being sold are: 1) 3 Loaves of Other bread and 3 baguettes. (Only one way of this happening) 2) 2 Loaves of Other bread and 4 baguettes.(2 ways of this happening since O1,O2 or O2,O2 assuming the other) 3) 1 Loaf of Other bread and 5 baguettes.(2 ways of this happening O1 or O2) 4) 6 baguettes.(Only one way of this happening)
So the probability is 2 favorable outcomes by 6 total outcomes = 1/3.
So the required piece of information is whether the 3 loaves of bread other than the baguettes are indistinguishable or not.
My entire solution relies on this below point that:
We are selecting a subset of indistinguishable objects from a set indistinguishable objects. There is only one way of doing this. The order in which the objects are selected does not matter since they are indistinguishable objects. The problem is like selecting 6 letters(and the order of this 6 letter is not relevant) from a set of letters which is {B,B,B,B,B,B,B,O,O,O}.
BBBBOO is the same as OBBOBB since at the end of the day 4 baguettes and 2 others got sold.
I hope i'm making sense. Please let me know why this approach is wrong. What assumption on my part is incorrect? I followed the same logic as go0ose. Why can't we assume that loaves of bread are the same? Should I look for explicit statement that items are the same? In case they are identical the answer is very simply. Imagine # of combination: 1) 6 B + 0 nonB 2) 5 B + 1 nonB 3) 4 B + 2 nonB 4) 3 B + 3 nonB 1/4!











