The probability that a visitor at the mall buys a pack of : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 14:30

### 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

# The probability that a visitor at the mall buys a pack of

Author Message
TAGS:

### Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2756
Location: New York City
Followers: 11

Kudos [?]: 857 [0], given: 4

The probability that a visitor at the mall buys a pack of [#permalink]

### Show Tags

13 Nov 2007, 08:20
8
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

62% (02:24) correct 38% (01:52) wrong based on 51 sessions

### HideShow timer Statistics

The probability that a visitor at the mall buys a pack of candy is 0.3. If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

[Reveal] Spoiler:
$$\frac{6!}{4!*2!}*0.3^4*0.7^2$$

Last edited by Bunuel on 24 Feb 2012, 12:21, edited 4 times in total.
Senior Manager
Joined: 06 Mar 2006
Posts: 492
Followers: 7

Kudos [?]: 179 [0], given: 1

### Show Tags

13 Nov 2007, 08:39
6 people and probability of buy candy 0.3

probability of exact 4 buys candy will be 0.3*0.3*0.3*0.3*0.7*0.7=0.003969
CEO
Joined: 21 Jan 2007
Posts: 2756
Location: New York City
Followers: 11

Kudos [?]: 857 [2] , given: 4

### Show Tags

13 Nov 2007, 08:42
2
KUDOS
eileen1017 wrote:
6 people and probability of buy candy 0.3

probability of exact 4 buys candy will be 0.3*0.3*0.3*0.3*0.7*0.7=0.003969

why isnt it 0.3*0.3*0.3*0.3*0.7*0.7 x 6C4
Senior Manager
Joined: 06 Mar 2006
Posts: 492
Followers: 7

Kudos [?]: 179 [0], given: 1

### Show Tags

13 Nov 2007, 09:06
Ok, this is just like toss the coin three times what is the probability that exact two will land on tail. Since each coin has 0.5 to be head and 0.5 to be tail. The probability will be 0.5(for land on tail)*0.5(for land on tail)*0.5( for land on head). This question has the same logic.

Never mind. My example is wrong. The probability of the coin will be 3/8.

Last edited by nevergiveup on 13 Nov 2007, 09:29, edited 1 time in total.
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 148 [0], given: 2

### Show Tags

13 Nov 2007, 09:09
i would have thought that it will be 6C4 , which gives the number of ways 4 people out of 6 can buy candy.

Then, take that number and multiply it by 0.3
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 68

Kudos [?]: 735 [0], given: 19

### Show Tags

13 Nov 2007, 22:31
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3.
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

p = 0.3^4 0.7^2

it cannot be multiples by 6c4 because the visitors are not going to have 4 buy andf 2 not-buy 6c4 times. we know for sure that it is going to happen only once.

therefore, p = 0.3^4 0.7^2 is correct.
Manager
Joined: 11 Nov 2007
Posts: 60
Followers: 1

Kudos [?]: 6 [2] , given: 0

### Show Tags

14 Nov 2007, 01:36
2
KUDOS
Here is how i approached it:

1. What is a prob of a simple sequence where the first 4 people were to buy so (0.3)^4 * (.7)^2

2. But this is also a combo problem because out of 6 people, any groups of 4 can be chosen to buy--not just the first 4. In another words, there are 6C4 ways to rearrange (0.3)(0.3)(0.3)(0.3)(0.7)(0.7).

So, you have to multiply by 6C4.
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 549

Kudos [?]: 3563 [0], given: 360

### Show Tags

23 Nov 2007, 23:09
a few theory statements....

p,q -probability, where p+q=1.
n - number of events.

Thus,
1^n=(p+q)^n=C(0,n)*p^n+C(1,n)*p^n*q+...+C(m,n)*p^(n-m)*q^m+...+C(n,n)*q^n

in our case:

P=C(m,n)*p^(n-m)*q^m, where m=4, n=6, p=0.7, q=0.3

Therefore,
C(4,6)*0.3^4*0.7^2
Intern
Joined: 13 Jan 2008
Posts: 24
Followers: 0

Kudos [?]: 36 [0], given: 0

### Show Tags

27 Jan 2008, 11:24
CaspAreaGuy wrote:
could it be (1-0.7*.03^5)*6?

Why have you subtracted the .07 from 1? That would simply make your equation (.3^6)*6. I was thinking (.3^5*.7)*6

???
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 422 [0], given: 0

### Show Tags

27 Jan 2008, 22:21
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

I also came across the same dilema here to multiply by 6C4 or not.

3/10*3/10*3/10*3/10*7/10*7/10

Now obvs we could have 6!/4!2!

Is this a problem you made up or is there an OA? Id like to see whether this is the correct approach. I went for multiplying by 6C4.

B/c there is no restriction on the order here. So we can have several different orders.
Senior Manager
Joined: 26 Jan 2008
Posts: 267
Followers: 4

Kudos [?]: 101 [1] , given: 16

### Show Tags

29 Jan 2008, 13:53
1
KUDOS
bmwhype2 wrote:
What is the P(at least one will not buy)?

Why is this not 1 - (0.3^6)?

Reason:
The probability that atleast one will not buy = 1 - (probability that everyone buys)
where (probability that everyone buys) = 0.3^6

(0.3^5) * (0.7) * 6 represents the probability that EXACTLY one person does not buy
_________________
Manager
Joined: 18 May 2007
Posts: 56
Followers: 1

Kudos [?]: 21 [1] , given: 0

### Show Tags

29 Jan 2008, 16:39
1
KUDOS
To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?
Senior Manager
Joined: 26 Jan 2008
Posts: 267
Followers: 4

Kudos [?]: 101 [1] , given: 16

### Show Tags

29 Jan 2008, 16:43
1
KUDOS
bhatia_ash2002 wrote:
To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?

0.7^6 is the probability that no one will buy
1 - (0.7^6) is hence the probability that atleast one person will buy, as opposed to the probability that atleast one person will not buy (as required by the question)
_________________
Manager
Joined: 18 Jul 2009
Posts: 169
Location: India
Schools: South Asian B-schools
Followers: 2

Kudos [?]: 98 [0], given: 37

### Show Tags

25 Jul 2009, 10:06
out of 4 visitors => 4 buys & 2 dont buy => 0.3^4 x 0.7^2 => 0.3969%
_________________

Bhushan S.
If you like my post....Consider it for Kudos

Intern
Joined: 26 Jul 2009
Posts: 17
Followers: 0

Kudos [?]: 7 [3] , given: 8

### Show Tags

21 Sep 2009, 00:55
3
KUDOS
Here is the brutal way of verifying the result ....

I think answer should be 6C4*(0.3)^4*(o.7)^2

EXAMPLE : Suppose we have 3 visitors ....
then complete tree would be like :

0.3*0.3*0.3 = 0.027 [ when all r gng for shopping]
0.3*0.3*0.7 = 0.063 [ 1&2 goes for shopping]
0.3*0.7*0.3 = 0.063 [ 1&3 goes for shopping]
0.7*0.3*0.3 = 0.063 [ 2&3 goes for shopping]
0.3*0.7*0.7 = 0.147 [ only 1 goes for shopping]
0.7*0.3*0.7 = 0.147 [ only 2 goes for shopping]
0.7*0.7*0.3 = 0.147 [ only 3 goes for shopping]
0.7*0.7*0.7 = 0.343 [ no one goes]

Total = 1

clearly If multiply by nCr ... only then we can get the correct answer

Consider for Kudos ....if u like it
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 144 [0], given: 3

### Show Tags

27 Sep 2009, 01:35
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

Soln:
6C4 * (.3)^4 * (.7)^2
Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 378 [0], given: 47

### Show Tags

14 Feb 2010, 09:03
bmwhype2 wrote:
the prob that a visitor at the mall buys a pack of candy is .3
If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?

6c4 x 0.3^4 x 0.7^2 = 15 x 0.0081 x 0.49 = 0.059
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Intern
Joined: 17 Jan 2010
Posts: 29
Schools: SSE, LSE
Followers: 0

Kudos [?]: 4 [0], given: 3

### Show Tags

04 Mar 2010, 04:07
incognito1 wrote:
bhatia_ash2002 wrote:
To the question:

What is the P(at least one will not buy)?

Cant we just rephrase it as:

What is the P(no one will buy)?

Which can be solved as

1-0.7^6

Why is this not the correct approach and answer?

0.7^6 is the probability that no one will buy
1 - (0.7^6) is hence the probability that atleast one person will buy, as opposed to the probability that atleast one person will not buy (as required by the question)

You are right!! The question asks for the probability that at least 1 does not buy. That means that everything has to be considered from 1 buying nothing to all 6 people buying nothing. Therefore the right solution should be 1 - p(everybody buys) = 1 - 0.3^6

And if everybody is buying the order does not matter in this respect.
Senior Manager
Joined: 11 May 2011
Posts: 372
Location: US
Followers: 3

Kudos [?]: 95 [0], given: 46

### Show Tags

12 Sep 2011, 15:25
bmwhype2 wrote:
What is the P(at least one will not buy)?

= 1-0.7^6

Let me know.
Cheers!
_________________

-----------------------------------------------------------------------------------------
What you do TODAY is important because you're exchanging a day of your life for it!
-----------------------------------------------------------------------------------------

Intern
Joined: 24 Feb 2012
Posts: 33
Followers: 0

Kudos [?]: 15 [0], given: 18

Re: the prob that a visitor at the mall buys a pack of candy is [#permalink]

### Show Tags

24 Feb 2012, 11:48
3
This post was
BOOKMARKED
6 trials
4 successes w/ prob of each success = 3/10
2 failures w/ prob of each failure = 7/10

6C4 x (3/10)^4 x (7/10)^2
=0.0595
=approximately 6%
Re: the prob that a visitor at the mall buys a pack of candy is   [#permalink] 24 Feb 2012, 11:48

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
In a Mall, there is an offer on perfumes " Buy 2 Get 1 free", that is 1 08 Jan 2017, 06:07
1 At the wholesale store you can buy an 8-pack of hot dogs for \$1.55, a 7 22 Dec 2015, 20:36
1 The probability that a visitor at the mall buys a pack of candy is 30% 4 03 Apr 2011, 15:14
91 The probability that a visitor at the mall buys a pack of 19 20 Oct 2009, 05:44
30 The probability that a visitor at the mall buys a pack of 20 16 Nov 2007, 08:27
Display posts from previous: Sort by