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

16 Nov 2007, 09:27

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

52% (01:58) correct
48% (00:56) wrong based on 539 sessions

HideShow timer Statistics

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?

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.

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?

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

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.

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

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

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) --> 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.

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: \(b(x; n, P) = nCx * P^x * (1 - P)^{n - x}\)

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

Show Tags

11 Sep 2013, 09:53

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. _________________

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) --> 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.

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

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) --> 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.

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).

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) --> 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.

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

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

Show Tags

03 Jun 2015, 13:19

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. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

I’m a little delirious because I’m a little sleep deprived. But whatever. I have to write this post because... I’M IN! Funnily enough, I actually missed the acceptance phone...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...