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Re M0409
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15 Sep 2014, 23:22
Official Solution: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 visitors will buy a pack of candy? A. 0.343 B. 0.147 C. 0.189 D. 0.063 E. 0.027 The event when 2 out of 3 visitors Buy a pack of candy can occur in \(\frac{3!}{2!}=3\) ways: BBN, BNB, NBB (\(\frac{3!}{2!}=3\) is basically the # of permutations of 3 letters out of which 2 B's are identical). Now, each B has the probability of 0.3 and N has the probability of \(10.3=07\), so \(P(B=2)=\frac{3!}{2!}*0.3^2*0.7=0.189\). Or you can directly apply the formula: if the probability of a certain event is \(p\) (0.3 in our case), then the probability of it occurring \(k\) times (2 times in our case) in \(n\)time (3 in our case) sequence is: \(P = C^k_n*p^k*(1p)^{nk}\) \(P = C^k_n*p^k*(1p)^{nk}= C^2_3*(0.3)^2*(10.3)^{32}=\frac{3!}{2!}*0.3^2*0.7=0.189.\) Answer: C
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Re: M0409
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19 Mar 2015, 11:34
Why do we multiply times 3 and not times 6?
Indeed the arrangement 0,3 * 0,3 * 0,7 has 3 combinations if order doesnt matter, but as order matters it must be 3!/(32)! = 6. Because they are distinct visitors so 0,3(tom) * 0,3(matt) * 0,7(joe) != 0,3(matt) * 0,3(tom) * 0,7(joe)
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Re: M0409
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20 Mar 2015, 04:56
madmax1000 wrote: Why do we multiply times 3 and not times 6?
Indeed the arrangement 0,3 * 0,3 * 0,7 has 3 combinations if order doesnt matter, but as order matters it must be 3!/(32)! = 6. Because they are distinct visitors so 0,3(tom) * 0,3(matt) * 0,7(joe) != 0,3(matt) * 0,3(tom) * 0,7(joe)
Thanks! When there are 3 visitors out of which 2 buy, there can be 3 cases only: BBN, BNB, NBB.
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Re: M0409
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09 Jul 2015, 07:31
Need help, How to use a slot method for this question ,



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10 Jul 2015, 23:43
Many thankss This type of question always make me confuse, how to improve my understanding, do I need to remember the formula,hope to hear from you Many thankss



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Re: M0409
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10 Jul 2015, 23:49
Ans C In the simplest way : apple08 , Bunuel: P(Buying candy)=0.3 So, P(Not buying)=0.7 There are 3 people and we need to chose exactly 2 who buy it. So, 0.3*0.3*0.7 (Bcz other person would not bu has to be taken into consideration here) =0.063 We can do this is 3 ways, different selections. therefore, 0.063*3 =0.189 Please hit KUDOS if it helped you!!



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11 Jul 2015, 14:41
Many thanksss



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Re M0409
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09 Aug 2015, 06:46
I think this is a highquality question. There is a typing error.
10.3=0.7
not 07



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Re: M0409
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10 Dec 2015, 06:55
I would like to know why orders matter in this case. Thank you



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28 Apr 2016, 05:00
nattiya27 wrote: I would like to know why orders matter in this case. Thank you I doesn't its a combination question: The way of choosing the 2 that will buy out of 3: 3C2 = 3
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Re: M0409
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26 Jan 2017, 20:25
I have a question.
I calculated my answer by simply (0.3)(0.3)(0.7) = 0.063 and got the wrong answer.
I thought this was the way we calculate probabilities... Now, do we multiply by 3 because we are asked for the possibility that EXACTLY two people out of three buy candy?
If we were to be asked a question without the mentioning of EXACTLY, would it make any difference?
In what cases can we just multiply the possibility of outcomes and what cases do we have to multiply the result by the total possible outcomes, which in this case is 3?
Thank you!



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Re: M0409
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27 Jan 2017, 06:18
conneryeon001 wrote: I have a question.
I calculated my answer by simply (0.3)(0.3)(0.7) = 0.063 and got the wrong answer.
I thought this was the way we calculate probabilities... Now, do we multiply by 3 because we are asked for the possibility that EXACTLY two people out of three buy candy?
If we were to be asked a question without the mentioning of EXACTLY, would it make any difference?
In what cases can we just multiply the possibility of outcomes and what cases do we have to multiply the result by the total possible outcomes, which in this case is 3?
Thank you! We are multiplying by 3 because the case when two out of three buys candy can occur in three different ways: BBN (first buys, second buys, third does not buy), BNB, NBB. Check different approaches HERE and HERE. Combinatorics Made Easy!Theory on CombinationsDS questions on CombinationsPS questions on CombinationsTough and tricky questions on CombinationsProbability Made Easy!Theory on probability problemsData Sufficiency Questions on ProbabilityProblem Solving Questions on ProbabilityTough Probability QuestionsHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M0409
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30 Jul 2018, 07:24
Question does not say that order matters. So I just used: .3* .3 * .7 what is here which indicates that the order matters?? Am I missing something?? I am confused about how to determine that Order is of importance here? Please Help..
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