Problem Solving

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..

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??

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.

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.

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?

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...

Thanks for validating my approach Ian.