Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Sep 2014, 20:33

Close

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
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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
User avatar
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 10 [0], given: 0

The probability that a visitor at the mall buys a pack of [#permalink] New post 16 Nov 2007, 08:27
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

57% (01:56) correct 43% (00:53) wrong based on 234 sessions
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

A. .343
B. .147
C. .189
D. .063
E. .027
[Reveal] Spoiler: OA
Manager
Manager
avatar
Joined: 20 Jun 2007
Posts: 157
Followers: 1

Kudos [?]: 13 [0], given: 0

 [#permalink] New post 16 Nov 2007, 08:33
(c)

Probablity that two buy candy is:

(0.3)(0.3)(0.7) = 0.063


Three ways that this can happen
A+B
A+C
B+C

3 * 0.063 = 0.189
Manager
Manager
User avatar
Joined: 02 Aug 2007
Posts: 146
Followers: 1

Kudos [?]: 10 [0], given: 0

 [#permalink] New post 16 Nov 2007, 09:31
If the question asked for the probability that customer 1 and customer 3 bought candy, the answer would be:

(.3)(.7)(.3) = .063

Is this reasoning right? Similar to the question about rain on the first 2 days of the week given a probability for rain.
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 196 [0], given: 0

GMAT Tests User
 [#permalink] New post 22 Dec 2007, 06:23
CaspAreaGuy wrote:
Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain?


Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy.

This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy.

(.3)(.3)(.7)*3 = .189
(.3)(.3)(.3)*1 = .027

.189 + .027 = .216 probability of at least 2 people buying candy.
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1829
Location: New York
Followers: 26

Kudos [?]: 445 [0], given: 5

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 22 Aug 2008, 07:46
yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027


0.3*0.3*0.7 + 0.3*0.3*0.7 + 0.3*0.3*0.7
= 0.189
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 21 Apr 2008
Posts: 272
Location: Motortown
Followers: 0

Kudos [?]: 81 [1] , given: 0

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 04 Oct 2008, 10:22
1
This post received
KUDOS
•P(Buy) = 3/10, P(No Buy) = 7/10
•2 Yes, 1 No = 3/10*3/10*7/10 = 63/1000
•3 Possibilities = YYN + YNY + NYY = 3(63/1000) = 189/1000 = .189%
Manager
Manager
User avatar
Joined: 01 Jan 2008
Posts: 227
Schools: Booth, Stern, Haas
Followers: 1

Kudos [?]: 44 [0], given: 2

GMAT Tests User
Re: [#permalink] New post 14 Oct 2008, 21:47
eschn3am wrote:
CaspAreaGuy wrote:
Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain?


Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy.

This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy.

(.3)(.3)(.7)*3 = .189
(.3)(.3)(.3)*1 = .027

.189 + .027 = .216 probability of at least 2 people buying candy.

can someone explain why should we multiply by three and one,
thanks in advance
2 KUDOS received
SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1579
Followers: 12

Kudos [?]: 181 [2] , given: 0

GMAT Tests User
Re: Re: [#permalink] New post 14 Oct 2008, 23:37
2
This post received
KUDOS
kazakhb wrote:
eschn3am wrote:
CaspAreaGuy wrote:
Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain?


Yes, right now the question says exactly two. So we need to find the probability that EXACTLY 2 people buy candy. If it said at least two, then we need to find the probability that 2 OR 3 people bought candy.

This would result in the answer we have for 2 people buying candy PLUS the probability of all 3 people buying candy.

(.3)(.3)(.7)*3 = .189
(.3)(.3)(.3)*1 = .027

.189 + .027 = .216 probability of at least 2 people buying candy.

can someone explain why should we multiply by three and one,
thanks in advance



Three conditions in which two people can buy....12, 23 or 13.
Only one condition in which all three people can buy...123.

Hence, the first probability is multiplied by 3 whereas the second probability is multiplied by only 1.
Director
Director
User avatar
Joined: 25 Oct 2008
Posts: 611
Location: Kolkata,India
Followers: 9

Kudos [?]: 164 [0], given: 100

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 24 Jul 2009, 18:13
Since the three people are DISTINCT thts why the anser is .063x3=.189:)
_________________

countdown-beginshas-ended-85483-40.html#p649902

Intern
Intern
avatar
Joined: 20 Aug 2009
Posts: 4
Followers: 1

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

Re: PS Candy Probability [#permalink] New post 12 Sep 2009, 17:44
Could someone please let me know where I've messed up my calculation?

P(exactly 2 visitors buying candy)
= 1 - P(exactly 3 visitors buying candy) - P(exactly 1 visitor buying candy) - P(no visitors buying candy)
= 1 - (3/10)^3 - 3/10*7/10*7/10 - (7/10)^3
= 1 - 0.027 - 0.147 - 0.343
= 0.483

Many thanks!
1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 20 Mar 2008
Posts: 455
Followers: 1

Kudos [?]: 67 [1] , given: 5

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 12 Sep 2009, 18:15
1
This post received
KUDOS
pinktyke wrote:
Could someone please let me know where I've messed up my calculation?

P(exactly 2 visitors buying candy)
= 1 - P(exactly 3 visitors buying candy) - P(exactly 1 visitor buying candy) - P(no visitors buying candy)
= 1 - (3/10)^3 - 3/10*7/10*7/10 - (7/10)^3
= 1 - 0.027 - 0.147 - 0.343
= 0.483

Many thanks!


P(exactly 1 visitor buying candy) = 3 * 3/10*7/10*7/10 = .441 (Between A, B & C it could be A or B or C)

or, P(exactly 2 visitors buying candy) = 1 - 0.027 - 0.441 - 0.343 = .189
1 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 23 Jun 2009
Posts: 360
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Followers: 5

Kudos [?]: 95 [1] , given: 60

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 12 Sep 2009, 18:17
1
This post received
KUDOS
pinktyke wrote:
Could someone please let me know where I've messed up my calculation?

P(exactly 2 visitors buying candy)
= 1 - P(exactly 3 visitors buying candy) - P(exactly 1 visitor buying candy) - P(no visitors buying candy)
= 1 - (3/10)^3 - 3/10*7/10*7/10 - (7/10)^3
= 1 - 0.027 - 0.147 - 0.343
= 0.483

Many thanks!

The possibility that exactly 1 visitor buying candy is three times you calculated. This is because positioning. 100, 010, 001 ;)
1-0,027-0,147*3-0,343
=1-0,027-0,441-0,343
=1-0,811
=0,189 ;)
Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 77 [0], given: 3

GMAT Tests User
Re: PS Candy Probability [#permalink] New post 27 Sep 2009, 01:27
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027

Soln:

Probability that exactly two will buy is = P(First Buys, Second Buys,Third does not buy) + P(First buys, Second does not buy,Third buys) + P(First does not buy, Second Buys,Third Buys)
= (3/10 * 3/10 * 7/10) * 3
= .189
Expert Post
6 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25238
Followers: 3429

Kudos [?]: 25228 [6] , given: 2702

Re: PS Candy Probability [#permalink] New post 09 Sep 2010, 20:31
6
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027


Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.

Explanation:
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.

NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.

Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?

The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.

So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441

NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least ONE buys, means that buys exactly one OR exactly two OR exactly three:

P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657

P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)

P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)

P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.

BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:

P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.

3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?

P(B=2)=5!/2!3!*0.3^2*0.7^3

We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.

Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 13 Jan 2012
Posts: 42
Followers: 0

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

Re: The probability that a visitor at the mall buys a pack of [#permalink] New post 23 Feb 2012, 14:07
The binomial probability formula seems like overkill for this, but I like to use it when I can so I can remember how to use it...

This page explains the Binomial Probability formula: http://stattrek.com/lesson2/binomial.aspx
Quote:
Suppose a binomial experiment consists of n trials and results in x successes. If the probability of success on an individual trial is P, then the binomial probability is:
b(x; n, P) = nCx * P^x * (1 - P)^{n - x}


In this problem:
n=3
x=2
p=3/10

b(x;n,p) = 3C2 * (3/10)^2 * (7/10)
= {3 * 3 * 3 * 7} / 1000
= .189

An alternate approach:
S implies Success, F implies Failure
P(exactly two successes) = P (SSF) + P (SFS) + P (FSS)
= (3/10 * 3/10 * 7/10) + (3/10 * 7/10 * 3/10) + (7/10 * 3/10 * 3/10)
= 3 * (3 * 3 * 7 / 1000)
= .189
SVP
SVP
User avatar
Joined: 09 Sep 2013
Posts: 2240
Followers: 186

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

Premium Member
Re: The probability that a visitor at the mall buys a pack of [#permalink] New post 11 Sep 2013, 08:53
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Intern
Intern
avatar
Joined: 21 Oct 2012
Posts: 21
Location: United States
Concentration: Marketing, Operations
GMAT 1: 650 Q42 V36
GPA: 3.6
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 9 [0], given: 15

Re: PS Candy Probability [#permalink] New post 14 May 2014, 20:40
Bunuel wrote:
yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027


Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.

Explanation:
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.

NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.

Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?

The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.

So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441

NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least ONE buys, means that buys exactly one OR exactly two OR exactly three:

P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657

P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)

P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)

P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.

BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:

P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.

3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?

P(B=2)=5!/2!3!*0.3^2*0.7^3

We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.

Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153

Hope it helps.



You said that probabilty of atleast 1 = 1 - probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25238
Followers: 3429

Kudos [?]: 25228 [1] , given: 2702

Re: PS Candy Probability [#permalink] New post 15 May 2014, 00:37
1
This post received
KUDOS
Expert's post
havoc7860 wrote:
Bunuel wrote:
yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027


Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.

Explanation:
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.

NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.

Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?

The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.

So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441

NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least ONE buys, means that buys exactly one OR exactly two OR exactly three:

P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657

P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)

P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)

P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.

BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:

P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.

3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?

P(B=2)=5!/2!3!*0.3^2*0.7^3

We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.

Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153

Hope it helps.


You said that probabilty of atleast 1 = 1 - probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this


I guess you are talking about example #2.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least 1 visitor buys out of 3, means 1, 2, or all 3 visitors buy, so all the cases but when no-one buys (while at most 1 out of 3 means 0 or 1). Hence the probability that at least 1 visitor buys out of 3 = 1 - (the probability that no-one buys).

Does this make sense?
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25238
Followers: 3429

Kudos [?]: 25228 [1] , given: 2702

Re: PS Candy Probability [#permalink] New post 15 May 2014, 00:40
1
This post received
KUDOS
Expert's post
havoc7860 wrote:
Bunuel wrote:
yuefei wrote:
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

a. .343
b. .147
c. .189
d. .063
e. .027


Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
Answer: C.

Explanation:
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.

NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.

Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?

The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.

So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441

NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.

2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?

At least ONE buys, means that buys exactly one OR exactly two OR exactly three:

P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657

P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)

P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)

P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.

BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:

P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.

3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?

P(B=2)=5!/2!3!*0.3^2*0.7^3

We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.

Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153

Hope it helps.



You said that probabilty of atleast 1 = 1 - probabiliy of 0, but won't probability of atleast 1 = probability of atmost 1? im a little confused as to how probablity of atleast 1 = probability of 0. Please help me with this


Some "at least" probability questions to practice:
leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html
a-fair-coin-is-tossed-4-times-what-is-the-probability-of-131592.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-132074.html
a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html
a-shipment-of-8-tv-sets-contains-2-black-and-white-sets-and-53338.html
on-a-shelf-there-are-6-hardback-books-and-2-paperback-book-135122.html
in-a-group-with-800-people-136839.html
the-probability-of-a-man-hitting-a-bulls-eye-in-one-fire-is-136935.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-141790.html
the-probability-that-a-convenience-store-has-cans-of-iced-128689.html
triplets-adam-bruce-and-charlie-enter-a-triathlon-if-132688.html
a-manufacturer-is-using-glass-as-the-surface-144642.html
the-probability-is-1-2-that-a-certain-coin-will-turn-up-head-144730.html (OG13)
a-fair-coin-is-to-be-tossed-twice-and-an-integer-is-to-be-148779.html
in-a-game-one-player-throws-two-fair-six-sided-die-at-the-151956.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: PS Candy Probability   [#permalink] 15 May 2014, 00:40
    Similar topics Author Replies Last post
Similar
Topics:
28 Experts publish their posts in the topic The probability that a visitor at the mall buys a pack of study 17 20 Oct 2009, 05:44
The probability that a visitor at the mall buys a pack of sarzan 2 10 Sep 2008, 17:37
3 The probability that a visitor at the mall buys a pack of stingraybullray 4 14 Jun 2008, 08:18
The probability that a visitor at the mall buys a pack of Amit05 1 24 Nov 2007, 07:45
10 Experts publish their posts in the topic The probability that a visitor at the mall buys a pack of bmwhype2 21 13 Nov 2007, 08:20
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.