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M03-26

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New post 16 Sep 2014, 00:21
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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. \(\frac{4}{7}\)

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New post 16 Sep 2014, 00:21
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 non-baguettes, 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 non-baguettes,

\(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\).

Where:

\(C^4_7\) - # of ways to choose 4 baguettes out of 7;

\(C^2_3\) - # of ways to choose 2 non-baguettes out of 3;

\(C^6_{10}\) - total # of ways to choose 6 loaves out of 10.


Answer: D
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Re: M03-26  [#permalink]

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New post 01 Jul 2015, 18:03
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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 non-baguettes, 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 non-baguettes,

\(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\).

Where:

\(C^4_7\) - # of ways to choose 4 baguettes out of 7;

\(C^2_3\) - # of ways to choose 2 non-baguettes 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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New post 01 Jul 2015, 18:51
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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 non-baguettes, 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 non-baguettes,

\(P(b=4)=\frac{C^4_7*C^2_3}{C^6_{10}}=\frac{1}{2}\).

Where:

\(C^4_7\) - # of ways to choose 4 baguettes out of 7;

\(C^2_3\) - # of ways to choose 2 non-baguettes 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 non-baguettes out of 3 non-baguettes .
so we have to find in how many way can we choose 4 baguettes out of 7baguettes and 2 non-baguettes out of 3 non-baguettes= 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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New post 16 Oct 2015, 10:54
I didn't understood why the baguetes were considered different.
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New post 17 Oct 2015, 01:00
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New post 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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New post 02 Feb 2016, 09:38
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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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New post 25 May 2016, 21:35
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 k-combination of a set S is a subset of k distinct elements of S. If the set has n elements, the number of k-combinations 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) (S-for 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 outcome
3S + 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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New post 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 k-combination of a set S is a subset of k distinct elements of S. If the set has n elements, the number of k-combinations 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) (S-for 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 outcome
3S + 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 non-baguettes 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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New post 24 Aug 2016, 03:12
I think this is a high-quality question and I agree with explanation.
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New post 27 Feb 2019, 09:19
I think this is a high-quality 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
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New post 12 Mar 2019, 23:10
Hi Bunuel
Can't we solve this question using the binomial situation (i.e by applying the formula 6C4 * (7/10)^4 * 3/10^2)?
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New post 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
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Re: M03-26   [#permalink] 11 Aug 2019, 16:38
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