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m04 q9

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m04 q9 [#permalink] New post 20 Dec 2008, 18:09
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49% (01:42) correct 50% (00:48) wrong based on 341 sessions
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

(A) 0.343
(B) 0.147
(C) 0.189
(D) 0.063
(E) 0.027

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The explained answer is:

0.3 x 0.3 x 0.7 x 2C3 (combination) = 0.3 x 0.3 x 0.7 x 3 = 0.189

Why is it not just 0.3 x 0.3 x 0.7? What is the significance of 2C3?
[Reveal] Spoiler: OA
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Re: m04 q9 [#permalink] New post 21 Dec 2008, 13:37
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0,3 x 0,3 x 0,7
0,3 x 0,7 x 0,3
0,7 x 0,3 x 0,3

I hope you understand :lol:
There are 3 combinations you can make with 3 persons and that's the significance of times 3.
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M04 Q9 [#permalink] New post 19 Feb 2009, 14:50
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?

0.343
0.147
0.189
0.063
0.027

I am wondering why C is the correct answer. I keep solving .3*.3*.7 which is .063 but in the reason why its correct it says that I should multiply by 3!/2! which doesnt make sense. Can anyone help. Thanks
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Re: M4 Q9 [#permalink] New post 19 Feb 2009, 19:31
RuthlessCA 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?

0.343
0.147
0.189
0.063
0.027

I am wondering why C is the correct answer. I keep solving .3*.3*.7 which is .063 but in the reason why its correct it says that I should multiply by 3!/2! which doesnt make sense. Can anyone help. Thanks


you need to multiply 0.063 by 3 because there 3 ways 2 visitors can be chosen;

= 3c2 (0.3)^2 (1-0.3)^(3-2) = 3 (0.3^2) (0.7) = 0.189
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Re: M4 Q9 [#permalink] New post 11 Feb 2010, 19:24
GMAT TIGER wrote:

you need to multiply 0.063 by 3 because there 3 ways 2 visitors can be chosen;

= 3c2 (0.3)^2 (1-0.3)^(3-2) = 3 (0.3^2) (0.7) = 0.189


I have a doubt.

If the question were to state : "what is the probability that any two will buy a pack of candy?" then would the answer be (0.3 x 0.3 x 0.7) = 0.063?

Am still trying to get my head wrapped around this probability thingy.

Thanks!!
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Re: m04 q9 [#permalink] New post 31 Mar 2010, 18:39
C=candy bought
X= No candy

Possible scenarios:
C C X
C X C
X C C
Case I: 0.3x0.3x0.7
Probabilities will be added for all 3 scenarios(prob for each scenario is the same .3x.3x.7)
Total probability: 3x(.3x.3x.7) =.189
Alternate:
3C2 : selecting 2 out of 3 where order isn't important C1, C2 cant be differentiated.
Prob: 3C2 x .3x.3x.7 =.189
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Re: m04 q9 [#permalink] New post 22 Apr 2010, 10:59
would somone please post a reply to ra011's question. I would really like to know the answer, i get stumped by almost all of the probability questions as there always seems to be some new reason why we should do it differently. Please help.

thanks
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Re: m04 q9 [#permalink] New post 20 May 2010, 05:37
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The probability of K things to happen from n things whose probability of happening is p and not happening is q. Given by

nck (p^k)(q^(n-k))

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Re: M4 Q9 [#permalink] New post 21 May 2010, 05:30
ra011y wrote:
GMAT TIGER wrote:

you need to multiply 0.063 by 3 because there 3 ways 2 visitors can be chosen;

= 3c2 (0.3)^2 (1-0.3)^(3-2) = 3 (0.3^2) (0.7) = 0.189


I have a doubt.

If the question were to state : "what is the probability that any two will buy a pack of candy?" then would the answer be (0.3 x 0.3 x 0.7) = 0.063?

Am still trying to get my head wrapped around this probability thingy.

Thanks!!


Think about it, if the constraint says "any two", it is less restrictive than saying a specific 2 out of the 3 or even exactly 2 out of 3. So any two out of the three buying candy would be the same answer as the OA + the probability of all 3 buying candy. (Since even in this case, 2 of them do buy candy, we have no restriction on the 3rd)

i.e. 0.189 + 0.3*0.3*0.3 = 0.216

Correct me if I'm wrong though.

EDIT: 0.063 would have been the answer in the following case: A, B and C visit the mall. What is the probability that only A and B will buy candy, and C does not?
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Re: m04 q9 [#permalink] New post 20 Aug 2010, 00:57
Its C.

0.3 x 0.3 x 0.7 x 3
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Re: m04 q9 [#permalink] New post 22 Aug 2010, 04:39
A lot of people always make this same mistake. When it comes to probability combined with combination problems, you have to consider the no of combination it can be arranged in.
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Re: m04 q9 [#permalink] New post 21 Nov 2010, 06:16
The answer is still not clear. Could please someone try to explain it one more time?
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Re: m04 q9 [#permalink] New post 21 Nov 2010, 06:23
Expert's post
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Re: m04 q9 [#permalink] New post 24 May 2011, 03:31
Probability = 3C2 * (0.3)^2 (0.7)

= 3 * 9/100* 7/10

= 189/1000

= 0.189

Answer - C
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Re: m04 q9 [#permalink] New post 24 May 2011, 19:56
Probabilty : .3x.3x.7= .063
Now consecutive combinations should be multiplied: in this case for three persons

Ans : 3x(.3x.3x.7)=0.189
Ans: C
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Re: m04 q9 [#permalink] New post 25 May 2011, 06:51
Kinda of confusing for sure. I forgot to take in acount the 3 different combinatons of people.
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Re: m04 q9 [#permalink] New post 29 May 2011, 03:33
C (.3)(.3)(.7) + (.7)(.3)(.3) + (.3)(.7)(.3) = .189
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Re: m04 q9 [#permalink] New post 13 Sep 2012, 08:58
Use binomial theorem to solve it in under 30 seconds. nCx*(prob.of event occuring)^x*(prob of event not occuring)^n-x
3C2*(.3)^2*(.7)^1
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Re: m04 q9 [#permalink] New post 14 Sep 2012, 10:36
Here we go, the question stem asks about EXACT number of events. Hence, to get the exactly the probability of two visitors buy a pack of candy can be find using binomial probability, or Bernulli trials consisting following steps:

1. 3C2 - number of ways how exactly 2 visitors could buy a pack of candy. Here it is easy, 3 (like this A,B,C visitors and we have AB, AC and BC - 3 ways).
2. raising our beneficial probability, i.e. 30% in our case to the power ow 2-exactly 2 customers in our case, hence 0.3^2 = 0.09
3. raising our failure probability, i.e. 1-0.3=0.7 to the power of 3-2=1, failure non-beneficial events.

Eventually, we just multiply all aforesaid three values and get the required probability, P = 3*0.09*0.7=27*7/1000 = 189/1000 or 0.189

So, I bet the answer should be (C), according to the Bernulli trials, I just followed what that math genius stated.

Please, correct me if I went awry.

georgechanhc 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) 0.343
(B) 0.147
(C) 0.189
(D) 0.063
(E) 0.027

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The explained answer is:

0.3 x 0.3 x 0.7 x 2C3 (combination) = 0.3 x 0.3 x 0.7 x 3 = 0.189

Why is it not just 0.3 x 0.3 x 0.7? What is the significance of 2C3?

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Re: m04 q9 [#permalink] New post 28 May 2013, 08:41
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Probability of a person to buy is 0.3 and probability of a person not to buy is 1-0.3 = 0.7

There can be 3 cases :

First and second person buy ,second and third person buy and third and first person buys the candy.

For each of the above three cases we have :

0.3*0.3*0.7 + 0.7*0.3*0.3 + 0.3*0.7*0.3

=0.189

Hence (C)
Re: m04 q9   [#permalink] 28 May 2013, 08:41
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