A store buys 10 bottles of alcohol, including 7 bottles of : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 02:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A store buys 10 bottles of alcohol, including 7 bottles of

Author Message
TAGS:

### Hide Tags

Manager
Joined: 17 May 2007
Posts: 72
Followers: 2

Kudos [?]: 29 [6] , given: 0

A store buys 10 bottles of alcohol, including 7 bottles of [#permalink]

### Show Tags

04 Nov 2007, 13:41
6
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

63% (03:11) correct 37% (02:47) wrong based on 796 sessions

### HideShow timer Statistics

A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7
[Reveal] Spoiler: OA
Director
Joined: 11 Jun 2007
Posts: 931
Followers: 1

Kudos [?]: 175 [6] , given: 0

### Show Tags

04 Nov 2007, 13:55
6
KUDOS
2
This post was
BOOKMARKED
gowani wrote:
please explain...I'm new to combinations and need help setting it up. Thanks in advance.

A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

please confirm if i am correct

10 bottles - 7 whiskeys / 3 other

6 sold, disregard 4

probability: F/T

T:
total possible: 10C6 = 10!/6!4! = 30*7 = 210

F:
selling 4 whiskeys: 7C4 = 7!/4!3! = 35
the rest will be 2 so: 3C2 = 3!/2!1! = 3
multiplying the two together : 35*3 = 105

the probability: F/T = 105/210 = 1/2 D

edit sorry to confuse i just updated

Last edited by beckee529 on 04 Nov 2007, 14:13, edited 1 time in total.
Manager
Joined: 17 May 2007
Posts: 72
Followers: 2

Kudos [?]: 29 [0], given: 0

### Show Tags

04 Nov 2007, 14:06
yes, you are correct.

what do you mean by:

the rest will be 4 so: 3C2 = 3!/2!1! = 3

also, what does F and T mean?

Director
Joined: 11 Jun 2007
Posts: 931
Followers: 1

Kudos [?]: 175 [0], given: 0

### Show Tags

04 Nov 2007, 14:11
gowani wrote:
yes, you are correct.

what do you mean by:

the rest will be 4 so: 3C2 = 3!/2!1! = 3

also, what does F and T mean?

the stem tells us that there were 6 sold and the rest are non whiskey alcohol. the "rest" is 4 so we do not consider those. Since they ask about 4 bottles of whiskey sold from the 6, 6-4 = 2 is the number of combinations for the 3 non whiskey bottles

F = favorable
T = total possibilities

those are the heart and core of probability that you have to fundamentally know to understand probability

for instance, 1/2 probability for head/tail coin flip

i hope that makes sense, good luck!
Intern
Joined: 26 Nov 2007
Posts: 6
Followers: 0

Kudos [?]: 1 [1] , given: 0

### Show Tags

26 Nov 2007, 19:59
1
KUDOS
how did you quickly derive that 10!/6!4! = 30*7 ?
Intern
Joined: 25 Nov 2007
Posts: 38
Followers: 0

Kudos [?]: 11 [2] , given: 0

### Show Tags

01 Dec 2007, 13:04
2
KUDOS

However something is confusing me when I look at it second time.

The only way 6 bottles are sold is if, We sell

1. 6 whisky
2. 5 whisky 1 Non whisky
3. 4 whisky and 2 non whisky
4. 3 whisky and 3 non whisky.

so total possible way of selling 6 bottles = 4
so total possible number when exactly 4 whiskys are sold = 1

so probabilty = 1/4

what is wrong with this equation??
Manager
Joined: 28 Jul 2008
Posts: 100
Followers: 1

Kudos [?]: 9 [1] , given: 0

### Show Tags

21 Sep 2008, 10:01
1
KUDOS
This is a classic hypergeometric distribution problem.
VP
Joined: 30 Jun 2008
Posts: 1043
Followers: 14

Kudos [?]: 567 [0], given: 1

### Show Tags

21 Sep 2008, 22:41
oziozzie wrote:
This is a classic hypergeometric distribution problem.

Whats that ??

gowani wrote:
please explain...I'm new to combinations and need help setting it up. Thanks in advance.
A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.
A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

I agree with above explanations.

Total outcomes : 10C6
Favourable Outcomes : 7C4*3C2

Probability : 7C4*3C2/10C6 = 105/210 =1/2
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 143 [0], given: 3

### Show Tags

28 Sep 2009, 10:10
A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

Soln: Probability that 4 bottles of whisky will be sold is
= 7C4 * 3C2/10C6
= 1/2

Ans is D
Senior Manager
Joined: 03 Nov 2005
Posts: 396
Location: Chicago, IL
Followers: 3

Kudos [?]: 50 [0], given: 17

### Show Tags

01 Oct 2009, 21:50
Solution:

C(6,4) / C(10,6)
_________________

Hard work is the main determinant of success

Intern
Joined: 18 Aug 2010
Posts: 10
Followers: 0

Kudos [?]: 3 [0], given: 1

### Show Tags

15 Sep 2010, 09:55
Can someone explain to me why the following won't work?
i did binomial approach
P (of getting whiskey among the 10 bottles) = 7/10
P(of getting something else) = 3/10

so getting 4 whiskey out of 6:
C (6,4)*(7/10)^4*(3/10)^2

but this doesn't give me 1/2 for an answer?
help?
_________________

D Day is April 23rd, 2010
Be humble, be focused, and be calm!

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 956 [0], given: 25

### Show Tags

15 Sep 2010, 10:06
The binomial approach only works if the probability of outcomes of each successive event remains constant

This is not true here

After the first bottle is sold, the probability of whisky and non-whisky changes.

Posted from my mobile device
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7074

Kudos [?]: 93067 [9] , given: 10541

### Show Tags

15 Sep 2010, 10:26
9
KUDOS
Expert's post
9
This post was
BOOKMARKED
MBAwannabe10 wrote:
Can someone explain to me why the following won't work?
i did binomial approach
P (of getting whiskey among the 10 bottles) = 7/10
P(of getting something else) = 3/10

so getting 4 whiskey out of 6:
C (6,4)*(7/10)^4*(3/10)^2

but this doesn't give me 1/2 for an answer?
help?

We can solve the way you propose as well, with a little correction.

A store buys 10 bottles of alcohol, including 7 bottles of whiskeys. In the evening, 6 bottles of alcohol are sold one by one, and the rest is consumed by the personnel. What is the probability of selling 4 whiskeys among the 6 bottles? Assume that every bottle has an equal chance of being bought.
A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

We have 7 whiskey (W) and 3 non-whiskey bottles and want to find the probability of selling WWWWNN (4 whiskey bottles and 2 non-whiskey bottles).

$$P=\frac{6!}{4!2!}*(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}*\frac{4}{7})*(\frac{3}{6}*\frac{2}{5})=\frac{1}{2}$$.

We are multiplying by $$\frac{6!}{4!2!}$$ as scenario WWWWNN can occur in # of ways:
NNWWWW (first bottle sold was non-whiskey, second bottle sold was non-whiskey, third bottle sold was whiskey, ...);
NWNWWW;
NWWNWW;
...

Some # of combinations, which basically equals to # of permuations of 6 leeters WWWWNN out of which 4 W's and 2 N's are identical --> $$\frac{6!}{4!2!}$$.

Similar problems:
viewtopic.php?f=140&t=56812&hilit=+probability+occurring+times
viewtopic.php?f=140&t=88069&hilit=+probability+occurring+times
viewtopic.php?f=140&t=87673&hilit=+probability+occurring+times
ps-probability-teams-68186.html?hilit=occur%20permutations%20letters#p728696
probability-question-100222.html?hilit=occur%20permutations%20letters#p772756

Hope it's clear.
_________________
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 690
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 15

Kudos [?]: 147 [2] , given: 15

### Show Tags

15 Sep 2010, 16:21
2
KUDOS
I guess this is simply 7C4 3C2 / 10C6 = 1/2?
_________________

Consider kudos, they are good for health

Manager
Joined: 04 Aug 2010
Posts: 158
Followers: 2

Kudos [?]: 27 [0], given: 15

### Show Tags

21 Sep 2010, 17:18
combinations are killing me! Great explanation Beckee
Intern
Joined: 10 Oct 2010
Posts: 23
Location: Texas
Followers: 3

Kudos [?]: 18 [0], given: 1

### Show Tags

11 Oct 2010, 04:21
gowani wrote:
A store buys 10 bottles of alcohol,

Bag of 10 choices.

gowani wrote:
including 7 bottles of whiskeys.

Bag is made of 7 whiskey and 3 nonwhiskey.

gowani wrote:
In the evening, 6 bottles of alcohol are sold one by one

Combo box arrangement
(_)(_)(_)(_)(_)(_)/6!

gowani wrote:
and the rest is consumed by the personnel.

um...ok. drinkdrinkdrinkdrinkdrink

gowani wrote:
What is the probability of selling 4 whiskeys among the 6 bottles?

Probability Table: Create and work backwards.

# of Whiskey: Events
0:
1:
2:
3:
4:
5:
6:
-------------------------------------
Total =

*******************************************************

Total = 10 bottles pick 6 = 10C6 = 10*9*8*7*6*5/6*5*4*3*2*1 = 210

6 Whiskey: 7 whiskey pick 6 = 7C6 = 7 <---------don't need this, but good for practice.

5 Whiskey: 7 whiskey pick 5 AND 3 non whiskey pick 1 = 7C5 * 3C1 = 63 <----------don't need this either.

4 Whiskey: 7 whiskey pick 4 AND 3 non pick 2 = 7C3 * 3C2 = 105 <-----This is what we need. I'll finish the table for fun.

3 Whiskey: 7 whiskey pick 3 AND 3 non pick 3 = 7C3 * 3C3 = 35
OR sum the table total: 210 - (7+63+105) = 35

2 Whiskey: 7 whiskey pick 2 AND 3 non pick 4 = impossible = 0

1 Whiskey: also impossible = 0

0 Whiskey: also impossible = 0

*******************************************************

# of Whiskey: Events
0: 0
1: 0
2: 0
3: 35
4: 105
5: 63
6: 7
-------------------------------------
Total = 210

*******************************************************

Use the info in the table to answer any probability question.
P(Whiskey = 4) = 105/210 = 1/2
P(Whiskey = prime number) = (0+35+63)/210
P(Whiskey > 3) = (105+63+7)/210

gowani wrote:
Assume that every bottle has an equal chance of being bought.

A. 2/5
B. 3/5
C. 2/3
D. 1/2
E. 4/7

ANS: D
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7119
Location: Pune, India
Followers: 2132

Kudos [?]: 13633 [3] , given: 222

### Show Tags

25 Jun 2012, 21:25
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
Responding to a pm:

We cannot use Binomial here. Think 'flipping a coin' or 'with replacement' when you want to use binomial.
You are not replacing the bottles with identical bottles every time one of them is sold/consumed. Hence, replacement is not taking place here. The probability of selling a whiskey bottle changes after you sell one. Therefore, the probability does not stay at 7/10 so you cannot use binomial.

You have 7 W and 3 N.
So probability of selling 4 W and 2 N = (7/10)*(6/9)*(5/8)*(4/7)*(3/6)*(2/5)* 6!/4!*2!

Now, as for your question: "A box contains 100 bulbs out of which 10 are defective. A sample of 5 bulbs is drawn.The prob. that none is defective is ?"

You need to give the exact question. Did they mention whether they are replacing the bulbs after each draw? If the solution uses binomial, when you draw a bulb, you need to replace it and then draw another one. Otherwise, you cannot use binomial here.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 03 Sep 2012 Posts: 338 Location: United States Concentration: Healthcare, Strategy GMAT 1: 730 Q48 V42 GPA: 3.88 WE: Medicine and Health (Health Care) Followers: 16 Kudos [?]: 179 [0], given: 31 Re: A store buys 10 bottles of alcohol, including 7 bottles of [#permalink] ### Show Tags 24 Sep 2012, 22:55 Total Number of Alcohol Bottles = 10 Whiskey Bottles = 7 No of bottles sold by the Store = 6 To calculate the probability that 4 out of these 6 is whiskey we will use the following formula : - P (A) = Total No. of Cases favorable to the happening of A ____________________________________________ Total No of exhaustive equally likely cases So the denominator becomes , C (10,6) which = 210 To find the number of combinations we use : C (7,4) x C (3,2) ie Four out of the 7 whiskey bottles makes up the four bottles (of the 6 sold) and 02 of the remaining non whiskey bottles make up the remaining two ... This is equal to 105 .... Dividing to get the probability we get 105/210 = 1/2 (D) _________________ "When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas GMAT Club Legend Joined: 09 Sep 2013 Posts: 13435 Followers: 575 Kudos [?]: 163 [0], given: 0 Re: A store buys 10 bottles of alcohol, including 7 bottles of [#permalink] ### Show Tags 10 Oct 2013, 19:31 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 09 Nov 2013 Posts: 17 Location: United Arab Emirates Concentration: Operations, Technology Schools: MBS '16 (A) GPA: 3.4 WE: Engineering (Energy and Utilities) Followers: 0 Kudos [?]: 7 [0], given: 18 Re: A store buys 10 bottles of alcohol, including 7 bottles of [#permalink] ### Show Tags 12 Nov 2013, 10:28 One way I thought you can look at the question is... what is the probability of 4 whiskeys and 2 bottles of alcohol being sold from a collection of 10 bottles if one bottle is being sold at a time. Thus the probability would be as follows: (7/10)*(6/9)*(5/8)*(4/7)*(3/6)*(2/5) = 1/30 => this is the probability that 4 whiskeys and 2 alcohol bottles are sold out of 10 bottles. but bottles can be sold in any order, so considering all orders in which the bottle may be sold that => 6!/(4!*2!) = 15 multiplying both, we get 15/30 => 1/2 I do hope that this method is right and hope that somebody benefits from this solution. Re: A store buys 10 bottles of alcohol, including 7 bottles of [#permalink] 12 Nov 2013, 10:28 Go to page 1 2 Next [ 26 posts ] Similar topics Replies Last post Similar Topics: 2 A bottle contains a certain solution. In the bottled solution, the rat 2 05 Apr 2016, 05:44 10 John Purchased 1375 large bottles at$1.89 per bottle and 69 4 19 Feb 2014, 06:13
2 A bottle is 80% full. The liquid in the bottle consists of 60% guava 18 11 Jul 2011, 03:37
69 A convenience store currently stocks 48 bottles of mineral water. The 18 27 Jun 2007, 16:26
Bottle R contains 250 capsules and costs \$6.25. Bottle T 3 07 Jan 2008, 02:37
Display posts from previous: Sort by