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Question Stats:
33% (01:22) correct
66% (00:59) wrong based on 1 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 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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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  There are 3 combinations you can make with 3 persons and that's the significance of times 3.
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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?
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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CEO
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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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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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Senior Manager
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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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Intern
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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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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))
-H
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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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Manager
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Its C.
0.3 x 0.3 x 0.7 x 3
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Manager
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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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Senior Manager
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The answer is still not clear. Could please someone try to explain it one more time?
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Probability = 3C2 * (0.3)^2 (0.7) = 3 * 9/100* 7/10 = 189/1000 = 0.189 Answer - C
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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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Kinda of confusing for sure. I forgot to take in acount the 3 different combinatons of people.
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C (.3)(.3)(.7) + (.7)(.3)(.3) + (.3)(.7)(.3) = .189
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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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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 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? _________________
God loves the steadfast.
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