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

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New post 16 Nov 2007, 09:27
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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
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Re: PS Candy Probability  [#permalink]

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New post 09 Sep 2010, 21:31
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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.
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New post 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
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New post 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.
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New post 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.
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Re: PS Candy Probability  [#permalink]

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New post 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
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Re: PS Candy Probability  [#permalink]

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New post 04 Oct 2008, 11:22
1
1
•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%
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New post 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,
thanks in advance
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New post 15 Oct 2008, 00:37
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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.
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New post 24 Jul 2009, 19:13
Since the three people are DISTINCT thts why the anser is .063x3=.189:)
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New post 12 Sep 2009, 18:44
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!
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New post 12 Sep 2009, 19:15
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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
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New post 12 Sep 2009, 19:17
1
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 ;)
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New post 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:

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
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New post 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...

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
\(P(exactly two successes) = P (SSF) + P (SFS) + P (FSS)\)
\(= (3/10 * 3/10 * 7/10) + (3/10 * 7/10 * 3/10) + (7/10 * 3/10 * 3/10)\)
\(= 3 * (3 * 3 * 7 / 1000)\)
\(= .189\)
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Re: PS Candy Probability  [#permalink]

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New post 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
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
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New post 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
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?
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New post 15 May 2014, 01:40
1
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
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.
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Re: The probability that a visitor at the mall buys a pack of  [#permalink]

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New post 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.

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

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Re: The probability that a visitor at the mall buys a pack of   [#permalink] 15 Sep 2018, 09:36
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