Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

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

Expert replies
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..
Join the discussion
Source: — Problem Solving |

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??
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Join the discussion

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
Join the discussion

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
Join the discussion

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!

Things are not what they appear to be... nor are they otherwise
Join the discussion

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...
Join the discussion

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

ianstewartgmat.com
Join the discussion

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