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]

Show Tags

13 Nov 2007, 08:20

8

This post was BOOKMARKED

00:00

A

B

C

D

E

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?

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.

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).

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.

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)
_________________

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

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?

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|

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! -----------------------------------------------------------------------------------------

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...