M03-26

Official Solution:

A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?

A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)


The store had a total of 10 loaves of bread, consisting of 7 baguettes and 3 non-baguettes.

Out of the 6 loaves sold, we want to determine the probability that exactly 4 were baguettes and 2 were non-baguettes.

Where:


Answer: D
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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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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 \(1/2\)

I think this is a high-quality question and I agree with explanation.

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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This is the approach I initially used before guessing around the 2 min mark. Can you explain why you are multiplying 6!/4! by 2!?

Thanks naumyuk for a great alternative solution.

There needs to be parentheses around the (4!*2!), but in this fraction notation that we normally use, the parentheses are implied: \(\frac{6!}{4!2!}\) = \(\frac{6*5}{2}=15\)

This is the standard "6 pick 4" (equivalent to "6 pick 2") combinations formula, which Bunuel generally notates in this form: \(C^4_{6}\), where \(C^r_{n}=\frac{n!}{r!(n-r)!}\)

Conceptually, suppose we are picking 2 out of 6: \(C^2_{6}\). There are 6 options for the first slot, and then only 5 options for the 2nd slot. However, since the order doesn't matter, we must divide by 2! ---> picking A then B is the same as picking B then A, so we've double counted the number of combinations. So, instead of the full formula, potentially a quicker way to do these is to just write \(\frac{6*5}{2}\). As another example, \(C^3_{7}\) becomes \(\frac{7*6*5}{3!}\)

adamfunk, any further questions on this?

I think this is a high-quality question and I agree with explanation. Great problem.

VeritasKarishma Bunuel chetan2u

I went through all the replies on this post and still did not find it convincing so I thought I'd reach out for an explanation

According to the link below and also from other posts that I scoured on the internet, whenever you have to choose from identical objects, there is only way of selecting them. In the link below you would see the same example of selecting of 2 identical balls from 5. There is only one way to do this (according to the article)


https://doubleroot.in/lessons/permutati ... l-objects/


So why shouldn't the same apply to this problem ? I am really confused.


Really appreciate the help on this !

Thanks
Siddharth

SiddharthR - Here is the problem with Combinatorics - a principle applicable to a certain scenario may become non-applicable if you even slightly change the data.
What is applicable to probability in case of "n identical objects" and also in case of "n identical objects of one kind and n identical objects of another kind" may not be applicable to "n identical objects of one kind and m identical objects of another kind".

The link you posted talks about lot 1 and lot 2 both having 3 objects each. The probability of picking an object from either lot is the same.
What happens when lot 1 has 7 objects and lot 2 has 3 objects? Will the probability of picking an object from lot 1 be the same as the probability of picking an object from lot 2? No. Lot 2 has far fewer objects.

Similarly, think about it: If you have 7 Gs and 3 Rs, will the probability of picking 3 Gs and 3 Rs be the same as the probability of picking 4 Gs and 2 Rs? No. The probability of picking 4 Gs and 2 Rs will be higher because of more number of Gs available.

Hence, the best way to deal with these questions is to use combinations assuming the objects are distinct (as done by Bunuel ). Since we are talking about a probability, it will not matter whether the objects are identical or distinct.
Or you can use the concepts of probability directly as done by naumyuk above.

SiddharthR

If I read this question, I cannot say for sure that they are identical as it is not clearly mentioned anywhere.

But say this question says they are basically two groups of identical things. 7 of identical As and 3 of identical Bs.
and I want exact 4 As, I could surely do it as following:-

Total possible combinations for these 6 are
AAABBB
AAAABB
AAAAAB
AAAAAA

We are looking for AAAABB

So P =1/4

But rest assured, GMAT will never put you in that tight spot or as we call a Dharam sankat.
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I also tried solving this way but I couldn't move beyond 1/30 -- Could you please help me get a better sense of clarity on when the order matters and when it doesn't. Or an easy way of recognizing the difference between - simultaneous and non-simultaneous events. Does the order matter here because they are sold? I considered them 6 as a whole; even if I moved baguettes up and down in this calculation (7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5 = 1/30), they should get adjusted, say if I was considering non-baguettes first: (3/10 * 2/9 * 7/8 * 6/7 * 5/6 * 4/5 = 1/30). How do I look out for these (non)-adjustments?

Please help, Bunuel naumyuk VeritasKarishma. Thank you.

7/10 * 6/9 * 5/8 * 4/7 * 3/6 * 2/5
What is 7/10? The probability of successfully picking a baguette out of the 10 loaves. This is your first pick.
Next, you pick another baguette with the probability 6/9. and so on...

So 1/30 is the probability of picking BBBBOO. This is not the probability of picking, say OOBBBB.

Look at a simpler case. Say you have 3 balls, Red, Blue and Yellow.

You can pick 2 balls in 3C2 = 3 ways (RB, BY, RY)
So Probability of picking Red & Blue is 1/3.

Probability of picking Red on your first pick is 1/3. Probability of then picking Blue is 1/2.
So RB probability is (1/3)*(1/2) = 1/6
BR probability is also 1/6.

Total probability of Red and Blue is 1/6 + 1/6 = 1/3

I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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