Last visit was: 19 Nov 2025, 16:16 It is currently 19 Nov 2025, 16:16
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
605-655 Level|   Probability|      
User avatar
yuefei
User avatar
Joined: 02 Aug 2007
Last visit: 02 Oct 2009
Posts: 45
Own Kudos:
251
 [53]
Posts: 45
Kudos: 251
 [53]
The probability that a visitor at the mall will purchase a pack of Updated on: 22 Aug 2025, 09:26
Originally posted by yuefei on 16 Nov 2007, 09:27.
Last edited by Bunuel on 22 Aug 2025, 09:26, edited 1 time in total.
5
Kudos
Add Kudos
48
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

67% (01:23) correct 33% (01:20) wrong based on 1303 sessions

History

Date
Time
Result
Not Attempted Yet
The probability that a visitor at the mall will purchase a pack of candy is 30%. If there are three visitors at the mall today, what is the probability that exactly two of them will buy a pack of candy?

A. 0.343
B. 0.147
C. 0.189
D. 0.063
E. 0.027

Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [60]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [60]
The probability that a visitor at the mall will purchase a pack of 03 Apr 2011, 16:14
5
Kudos
Add Kudos
55
Bookmarks
Bookmark this Post
Official Solution:

The probability that a visitor at the mall will purchase a pack of candy is 30%. If there are three visitors at the mall today, what is the probability that exactly two of them will buy a pack of candy?

A. 0.343
B. 0.147
C. 0.189
D. 0.063
E. 0.027


When examining the event where exactly 2 out of 3 visitors purchase a pack of candy, there are \(\frac{3!}{2!} = 3\) possible scenarios: (Buy, Buy, Not Buy); (Buy, Not Buy, Buy); and (Not Buy, Buy, Buy). The calculation \(\frac{3!}{2!} = 3\) represents the number of permutations of the pattern BBN (Buy, Buy, Not Buy).

Furthermore, each "Buy" event has a probability of 0.3, while each "Not Buy" event has a probability of \(1 - 0.3 = 0.7\). Therefore, the probability of having exactly 2 out of 3 visitors buy a pack of candy is given by \(P(B=2) = \frac{3!}{2!} *0.3^2 *0.7 = 0.189\).


Answer: C
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [30]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [30]
The probability that a visitor at the mall will purchase a pack of 09 Sep 2010, 21:31
18
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
yuefei
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
Show more

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.
General Discussion
User avatar
Raffie
User avatar
Joined: 20 Jun 2007
Last visit: 22 Feb 2008
Posts: 58
Own Kudos:
81
 [12]
Posts: 58
Kudos: 81
 [12]
The probability that a visitor at the mall will purchase a pack of 16 Nov 2007, 09:33
8
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
(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
User avatar
yuefei
User avatar
Joined: 02 Aug 2007
Last visit: 02 Oct 2009
Posts: 45
Own Kudos:
251
Posts: 45
Kudos: 251
The probability that a visitor at the mall will purchase a pack of 16 Nov 2007, 10:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
eschn3am
User avatar
Joined: 12 Jul 2007
Last visit: 03 Apr 2017
Posts: 396
Own Kudos:
582
 [4]
Posts: 396
Kudos: 582
 [4]
The probability that a visitor at the mall will purchase a pack of 22 Dec 2007, 07:23
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
CaspAreaGuy
Guys, will the answer be different if the question asked for a probablity that at least two will buy candies? Can anyone please explain?
Show more


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.
User avatar
dreambeliever
User avatar
Joined: 24 Nov 2010
Last visit: 20 Jun 2013
Posts: 118
Own Kudos:
291
 [6]
Given Kudos: 7
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Duke '15 Ross '15
Schools: Duke '15 Ross '15
Posts: 118
Kudos: 291
 [6]
The probability that a visitor at the mall will purchase a pack of 03 Apr 2011, 16:45
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
__ __ __
.3 .3 .7

__ __ __
.3 .7 .3

__ __ __
.7 .3 .3

probability of exactly 3 people buying it is 3(.3*.3*.7) = .189

Ans C
User avatar
subhashghosh
User avatar
Retired Moderator
Joined: 16 Nov 2010
Last visit: 25 Jun 2024
Posts: 896
Own Kudos:
1,279
 [5]
Given Kudos: 43
Location: United States (IN)
Concentration: Strategy, Technology
Products:
My Reviews
Posts: 896
Kudos: 1,279
 [5]
The probability that a visitor at the mall will purchase a pack of 03 Apr 2011, 17:58
1
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
3C2 * (3/10)^2 * 7/10

= 3 * 9/100 * 7/10

= 189/1000

= 0.189

Answer - C

For such questions, there is the formula I've used above. Please check it in Walker's Probability tutorial.
avatar
havoc7860
avatar
Joined: 21 Oct 2012
Last visit: 12 Jan 2015
Posts: 27
Own Kudos:
67
Given Kudos: 19
Location: United States
Concentration: Marketing, Operations
GMAT 1: 650 Q44 V35
GMAT 2: 600 Q47 V26
GMAT 3: 660 Q43 V38
GPA: 3.6
WE:Information Technology (Computer Software)
GMAT 3: 660 Q43 V38
Posts: 27
Kudos: 67
The probability that a visitor at the mall will purchase a pack of 14 May 2014, 21:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
yuefei
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.
Show more


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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [1]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [1]
The probability that a visitor at the mall will purchase a pack of 15 May 2014, 01:37
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
havoc7860
Bunuel
yuefei
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
Show more

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?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [3]
The probability that a visitor at the mall will purchase a pack of 15 May 2014, 01:40
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
havoc7860
Bunuel
yuefei
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
Show more

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.
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,977
Own Kudos:
8,391
Given Kudos: 1,646
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 2,977
Kudos: 8,391
The probability that a visitor at the mall will purchase a pack of 02 Sep 2017, 07:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
yuefei
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
Show more

We need to determine the probability that two out of three visitors will buy a pack of candy:

P(Y-Y-N) = 0.3 x 0.3 x 0.7 = 0.063

Since there are 3 ways -- (Y-Y-N), (Y-N-Y), or (N-Y-Y) -- in which two of the three visitors can buy a pack of candy, the overall probability is 3 x 0.063 = 0.189.

Answer: C
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 19 Nov 2025
Posts: 4,844
Own Kudos:
8,945
 [1]
Given Kudos: 225
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,844
Kudos: 8,945
 [1]
The probability that a visitor at the mall will purchase a pack of 13 Oct 2019, 04:43
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
This is a fairly straightforward question on Probability. This tests you not only on your understanding of the basic concepts of Probability but also on your ability to apply P&C concepts.

The first thing you can do in such questions is to write down the respective probabilities and simplify them to a fraction. It’s always easier to deal with fractions than with decimals, since you can perform all the mathematical operations on fractions quite easily.

It’s given in the question that the probability of a visitor buying a candy is 30% i.e. \(\frac{3}{10}\). Therefore, we can deduce that the probability of a visitor not buying a candy will be 70% i.e. \(\frac{7}{10}\). These are the respective probabilities.

Of the three visitors to the mall, we want exactly two of them to buy candies. This means that the third visitor should not buy a candy. Since we have to select any 2 visitors out of 3 visitors, who would buy the candy, we can do this in \(3_C_2\) i.e. 3 ways.

In each of these 3 ways, probability that 2 persons buy a candy and 1 does not = \(\frac{3}{10} * \frac{3}{10} * \frac{7}{10}\) = \(\frac{63}{1000}\).

Therefore, the total probability = 3 * \(\frac{63}{1000}\) = \(\frac{189}{1000}\) = 0.189. The correct answer option is C.

Hope that helps!
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,589
Own Kudos:
1,079
Posts: 38,589
Kudos: 1,079
The probability that a visitor at the mall will purchase a pack of 15 Sep 2025, 06:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts