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probability.. got the right answer but did i use wr medthod?

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probability.. got the right answer but did i use wr medthod?

by what? » Fri Jun 10, 2011 12:24 am
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.

2/5
3/5
2/3
1/2
4/7


what i did was <probability of choosing 4 baguettes in a row>* (probability of choosing 2 normal loaves) =< 7/10*6/9*5/8*4/7 > * (3/6*2/5) = 1/30

number of ways to arrange the 3 loaves = 6C4= 15.

multiply 1/30 * 15 and you have 1/2.

explanation given

Find the probability by dividing the number of favorable outcomes by the number of total outcomes.

Favorable outcomes (combinations of 4 baguettes and 2 other loaves): 7c4 * 3c2 = 105

Total outcomes (combinations of 6 loaves out of 10): 10c6 = 210

The probability is = 105/210 = 1/2. same answer.. different method. Am i wrong..
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by cans » Fri Jun 10, 2011 5:53 am
7 B and 3 NB (we are given total 10)
4 B sold and 2NB sold
total possible ways = 10C6 = 210
with constraint: 7C4*3C2 = 35*3
probability = 35*3/210 = 1/2
in the first method, why did you multiply it by 6C4??
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by bubbliiiiiiii » Fri Jun 10, 2011 6:04 am
cans wrote: in the first method, why did you multiply it by 6C4??
He did so to rearrange the 4 elements i.e., 3 loaves of bread as three elements and the fourth one being the group of 7 Balequette breads.

@what?

I don't think you have used the wrong method but I would rather say a complicated method we could directly obtain the answer as cans have derived it because there are no restrictions on the sale of the type of bread. i.e., every loaf has an equal chance of selling.

Hope it clears your query.

Please correct incase I am wrong.
Regards,

Pranay
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by bubbliiiiiiii » Fri Jun 10, 2011 6:08 am
cans wrote: in the first method, why did you multiply it by 6C4??
He did so to rearrange the 4 elements i.e., 3 loaves of bread as three elements and the fourth one being the group of 7 Balequette breads.

@what?

I don't think you have used the wrong method but I would rather say a complicated method we could directly obtain the answer as cans have derived it because there are no restrictions on the sale of the type of bread. i.e., every loaf has an equal chance of selling.

Hope it clears your query.

Please correct incase I am wrong.
Regards,

Pranay
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by Frankenstein » Fri Jun 10, 2011 6:11 am
bubbliiiiiiii wrote:
cans wrote: in the first method, why did you multiply it by 6C4??
He did so to rearrange the 4 elements i.e., 3 loaves of bread as three elements and the fourth one being the group of 7 Balequette breads.

@what?

I don't think you have used the wrong method but I would rather say a complicated method we could directly obtain the answer as cans have derived it because there are no restrictions on the sale of the type of bread. i.e., every loaf has an equal chance of selling.

Hope it clears your query.

Please correct incase I am wrong.
Hi,
It is a selection problem. I don't understand what you mean by rearranging. That too arranging 4 elements in 6C4 ways?
Cheers!

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by what? » Sat Jun 11, 2011 10:34 pm
so maybe it was because i had done too many probability problems that day, but i think that method sort of still makes sense.

to answer the question as to why i multiplied by 6C4, it is because when I calculated 1/30 , that was the probability of one possible arrangement of of 4 baguettes(B) and 2 normal loaves(L)
i.e. probability of BBBBLL ... but the arrangement BBBLLB is also and acceptable outcome .. and there are others too. how many arrangements are there? that will depend on how you arrange the 4 B's OR the 2L throughout that arrangement , i.e. 6C4 (if arranging B) or = 6C2 ( if arranging L)

note we use combination here because the identify of each individual loaf does not matter...
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by Ian Stewart » Sun Jun 12, 2011 12:07 am
Yes, what?, your method makes perfect sense. I find it just as fast as the other method suggested above, so if it makes more sense to you, there's no reason not to use it. I actually find your approach easier to generalize to other (related) probability problems. For example, if I see a question like:

The probability of rain on any given day in City X is 0.7. What is the probability it rains exactly 3 times in a certain 5 day period in City X?

then I'll first figure out the probability of one specific sequence, say Rain, Rain, Rain, Sun, Sun, which would be (0.7)(0.7)(0.7)(0.3)(0.3) = (0.7)^3(0.3)^2. Every such sequence is equally likely, and since there are 5C3 = 10 ways to choose which 3 of the 5 days will be rainy, there are 10 such sequences, so we'd just multiply (0.7)^3(0.3)^2 by 10 to get the answer.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by what? » Sun Jun 12, 2011 11:24 am
Thanks for validating my approach Ian.
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