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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
The probability that a visitor at the mall buys a pack of [#permalink]
13 Nov 2007, 08:20
3
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
68% (02:24) correct
32% (01:35) wrong based on 30 sessions
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?
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.
Re: probability - candy [#permalink]
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.
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).
Re: probability - candy [#permalink]
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.
Re: probability - candy [#permalink]
29 Jan 2008, 16:43
1
This post received 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) _________________
Re: probability - candy [#permalink]
21 Sep 2009, 00:55
3
This post received 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
Re: probability - candy [#permalink]
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?
Re: probability - candy [#permalink]
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!!
|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
Re: probability - candy [#permalink]
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.
P(at least one will not buy) = 1 - P(Everybody Buys) = 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! -----------------------------------------------------------------------------------------
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...