GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Sep 2018, 06:53

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

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

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

Author Message
TAGS:

Hide Tags

Manager
Joined: 02 Aug 2007
Posts: 142
The probability that a visitor at the mall buys a pack of  [#permalink]

Show Tags

16 Nov 2007, 09:27
2
16
00:00

Difficulty:

55% (hard)

Question Stats:

54% (01:00) correct 46% (00:59) wrong based on 759 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?

A. .343
B. .147
C. .189
D. .063
E. .027
Math Expert
Joined: 02 Sep 2009
Posts: 49300

Show Tags

09 Sep 2010, 21:31
9
3
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

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
Manager
Joined: 20 Jun 2007
Posts: 151

Show Tags

16 Nov 2007, 09:33
1
(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
Joined: 02 Aug 2007
Posts: 142

Show Tags

16 Nov 2007, 10:31
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
Joined: 12 Jul 2007
Posts: 849

Show Tags

22 Dec 2007, 07:23
1
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
Joined: 07 Nov 2007
Posts: 1708
Location: New York

Show Tags

22 Aug 2008, 08:46
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
_________________

Smiling wins more friends than frowning

Senior Manager
Joined: 21 Apr 2008
Posts: 263
Location: Motortown

Show Tags

04 Oct 2008, 11:22
1
1
•2 Yes, 1 No = 3/10*3/10*7/10 = 63/1000
•3 Possibilities = YYN + YNY + NYY = 3(63/1000) = 189/1000 = .189%
Manager
Joined: 01 Jan 2008
Posts: 216
Schools: Booth, Stern, Haas

Show Tags

14 Oct 2008, 22:47
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,
VP
Joined: 17 Jun 2008
Posts: 1463

Show Tags

15 Oct 2008, 00:37
2
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,

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
Joined: 25 Oct 2008
Posts: 541
Location: Kolkata,India

Show Tags

24 Jul 2009, 19:13
Since the three people are DISTINCT thts why the anser is .063x3=.189:)
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Intern
Joined: 20 Aug 2009
Posts: 4

Show Tags

12 Sep 2009, 18:44
Could someone please let me know where I've messed up my calculation?

= 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!
Senior Manager
Joined: 20 Mar 2008
Posts: 433

Show Tags

12 Sep 2009, 19:15
2
pinktyke wrote:
Could someone please let me know where I've messed up my calculation?

= 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
Senior Manager
Joined: 23 Jun 2009
Posts: 353
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago

Show Tags

12 Sep 2009, 19:17
1
pinktyke wrote:
Could someone please let me know where I've messed up my calculation?

= 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
Joined: 27 Oct 2008
Posts: 180

Show Tags

27 Sep 2009, 02:27
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:

= (3/10 * 3/10 * 7/10) * 3
= .189
Intern
Joined: 13 Jan 2012
Posts: 39
Re: The probability that a visitor at the mall buys a pack of  [#permalink]

Show Tags

23 Feb 2012, 15:07
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...

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
$$P(exactly two successes) = P (SSF) + P (SFS) + P (FSS)$$
$$= (3/10 * 3/10 * 7/10) + (3/10 * 7/10 * 3/10) + (7/10 * 3/10 * 3/10)$$
$$= 3 * (3 * 3 * 7 / 1000)$$
$$= .189$$
Intern
Joined: 21 Oct 2012
Posts: 34
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)

Show Tags

14 May 2014, 21:40
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

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
Math Expert
Joined: 02 Sep 2009
Posts: 49300

Show Tags

15 May 2014, 01:37
1
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

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?
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49300

Show Tags

15 May 2014, 01:40
1
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

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
a-manufacturer-is-using-glass-as-the-surface-144642.html
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.
_________________
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835
Re: The probability that a visitor at the mall buys a pack of  [#permalink]

Show Tags

02 Sep 2017, 07:21
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

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.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Non-Human User
Joined: 09 Sep 2013
Posts: 8121
Re: The probability that a visitor at the mall buys a pack of  [#permalink]

Show Tags

15 Sep 2018, 09:36
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.
_________________
Re: The probability that a visitor at the mall buys a pack of &nbs [#permalink] 15 Sep 2018, 09:36
Display posts from previous: Sort by

Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.