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Re M1014 [#permalink]
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16 Sep 2014, 00:42
Official Solution:Set \(S\) consists of all prime integers less than 10. If a number is selected from set \(S\) at random and then another number, not necessarily different, is selected from set \(S\) at random, what is the probability that the sum of these numbers is odd? A. \(\frac{1}{8}\) B. \(\frac{1}{6}\) C. \(\frac{3}{8}\) D. \(\frac{1}{2}\) E. \(\frac{5}{8}\) Set \(S =\{2, 3, 5, 7\}\). The question "what is the probability that the sum of these numbers is odd?" is equivalent to the question "what is the probability that one of these numbers is 2 while the other is not?". \(P(\text{the sum is odd}) = (P(\text{the first number is 2}) * P(\text{the second number is not 2})) +\) \(+ (P(\text{the first number is not 2}) * P(\text{the second number is 2})) =\) \(= \frac{1}{4}*\frac{3}{4} + \frac{3}{4}*\frac{1}{4} = \frac{6}{16} = \frac{3}{8}\). Answer: C
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Re: M1014 [#permalink]
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05 Dec 2014, 21:54
I quickly listed out the set of numbers (2,2) (2,3) (2,5) (2,7) (3,3) (3,5) (3,7) (5,5) (5,7) and (7,7) and then I counted the only odd pair which left me with a probability of 3/10. Why is this method incorrect?



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Re: M1014 [#permalink]
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06 Dec 2014, 06:24
rsamant wrote: I quickly listed out the set of numbers (2,2) (2,3) (2,5) (2,7) (3,3) (3,5) (3,7) (5,5) (5,7) and (7,7) and then I counted the only odd pair which left me with a probability of 3/10. Why is this method incorrect? There are more possibilities when picking two numbers: (2,2) (2,3) (3,2) (2,5) (5,2) (2,7) (7,2) (3,3) (3,5) (5,3) (3,7) (7,3) (5,5) (5,7) (7,5) (7,7) P = 6/16 = 3/8.
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This question is basically asking If two numbers are randomly selected from the set {2,3,5,7}, what is the probability that EXACTLY one of them is 2?(and suddenly it becomes a sub600 level ) P(Exactly one 2) = [P(2 in first pick) AND P(not 2 in second pick)] OR [P(not 2 in first pick) and P(2 in second pick)]\(=> \frac{1}{4} * \frac{3}{4} + \frac{3}{4} * \frac{1}{4}\) \(=> 2* \frac{3}{16}\) \(=> \frac{3}{8}\) Option C
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Re: M1014 [#permalink]
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03 Aug 2016, 00:05
I struggle to know when order matters. Why is the answer not simply 1/4*3/4?
Why is picking 2 then 3 or 5 or 7 different from picking picking 3 or 5 or 7 then 2?



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if a number is drawn then remaining numbers are 3 out of 4
total cases  4C2*2 ( first selected 02 numbers out of 4 and again as order matters, multiply by 2)= 12
now favorable cases where 2 is always selected so the combination is (2,3), (2,5), (2,7) again as order matters (3,2), (5,2) (7,2) also possible so total cases = 03+03= 06
probability = 06/12= 1/2
now if you say that if a number is withdrawn and you can draw the same number again for the second chance, hard to understand, until unless it is mentioned that the same number is available in the pool all the time



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Re: M1014 [#permalink]
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27 Sep 2016, 14:48
how can i calculate the total outcomes for this questions? i got the p(e)=6 but struggled to calculate the total outcomes through a formula.



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Re: M1014 [#permalink]
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13 Jan 2017, 07:24
The sum to be odd, the numbers selected must be either {2,3}, {2,5} or {2,7}.
The probability to drawn the numbers 2 and 3 is 2*\(\frac{1}{4}\)*\(\frac{1}{4}\) (We multiply by 2 because the numbers drawn could be either 2 and 3 OR 3 and 2 => so there are 2 cases AB, BA) The probability is the same for the numbers {2,5} and {2,7}.
So, the probability that the sum of the 4 numbers is odd is: 2*\(\frac{1}{4}\)*\(\frac{1}{4}\)+2*\(\frac{1}{4}\)*\(\frac{1}{4}\)+2*\(\frac{1}{4}\)*\(\frac{1}{4}\)=\(\frac{3}{8}\)



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Re: M1014 [#permalink]
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16 Jan 2017, 11:53
I used the below Proba(sum is odd) = 1  Proba(sum is even) The set is made of {2;3;5;7}. Number of all possible events is 4^2, which is 16 The sum is even if and only if you : pick twice an even number : there is only {2;2}, ie. 1 possibility or pick 2 odd numbers out of the 3 in the set: 3^2 ie. 9 possiblities. Therefore, Proba(sum is even)= (1+9)/16 = 10/16 and hence Proba(sum is odd)= 6/16 = 3/8 Done!!!
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Re: M1014 [#permalink]
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26 Mar 2017, 04:56
The question says "not necessarily different". Can we safely assume that the first number will be pot in the box again?



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Re: M1014 [#permalink]
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26 Mar 2017, 05:00



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Re: M1014 [#permalink]
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26 Mar 2017, 09:46
Bunuel wrote: dyg wrote: The question says "not necessarily different". Can we safely assume that the first number will be pot in the box again? It implies that numbers can be repeated. That's why the denominator in the solution is 4 for the second pick too. Yes I got it but I actually wanted to ask is "is it always OK to repeat numbers/colors etc whenever we see "not necessarily different"? Thanks.



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Re: M1014 [#permalink]
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26 Mar 2017, 22:49



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Re: M1014 [#permalink]
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01 Jul 2017, 03:13
I solved in the different way though which I figured after I got it wrong.
Since the order of the selection matters, we should not use combination formula, use permutation.
S = (2,3,5,7)
Picking two numbers from the set is 4P2= 12.
but the stem specifically says that choosing the same number is also possible which is (2,2)(3,3)(5,5)(7,7)
So total number of ways = 12+4 = 16 ways.
The rest is easy. Sum of two numbers is ODD. We know that Odd+Even = Odd (Since 2 is the only even)
(2, 3 or 5 or 7) or (3 or 5 or 7 , 2). We know that OR means addition. So total 6 ways.
6/16 = 3/8



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Re M1014 [#permalink]
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03 Jul 2017, 00:39
I think this is a poorquality question and I don't agree with the explanation. i want to understand. what do you mean by "not necessarily different? that means Set 'S' can have repetition of prime numbers. also, what if set S has (2,2,2,2,3,5,5,7,7) or any combinations of prime numbers less than 10? i marked the answer as a guess.
please clarify



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Re: M1014 [#permalink]
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03 Jul 2017, 00:46
Sushilait84 wrote: I think this is a poorquality question and I don't agree with the explanation. i want to understand. what do you mean by "not necessarily different? that means Set 'S' can have repetition of prime numbers. also, what if set S has (2,2,2,2,3,5,5,7,7) or any combinations of prime numbers less than 10? i marked the answer as a guess.
please clarify The question is mathematically very precise. It implies that numbers can be repeated. That's why the denominator in the solution is 4 for the second pick too. For example, you can choose 2 and 2 OR 7 and 7. All possible combinations are given here: https://gmatclub.com/forum/m10183859.html#p1452001
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Re: M1014 [#permalink]
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17 Oct 2017, 23:06
Hello BunuelI think this is a highquality question and I agree with the solution. I have the same conceptual gap as Demba below. Can you please help? Usually, for questions like these, I find the total possibilities first and then the favourable outcomes. To do this, its necessary to know if the order of outcome matters. Clearly, in this question order matters. (2,3) is different from (3,2). How do we understand after reading the question that order matters? Is it because the question says 'not necessarily different' which implies repetition is allowed? I was comparing this solution with this one  https://gmatclub.com/forum/m20184246.html#p1859217. In this solution that I pasted, Order does not matter. So, we wanted to know how to decide after reading a question if order matters or no. I hope I was able to communicate what I intend to say. Thanking you in advance! Demba wrote: I struggle to know when order matters. Why is the answer not simply 1/4*3/4?
Why is picking 2 then 3 or 5 or 7 different from picking picking 3 or 5 or 7 then 2?
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Re: M1014 [#permalink]
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31 Dec 2017, 09:17
P (EO) = 1/4 *3/4* 2! = 3/8 If you have any confusion with this method let me know.



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Re: M1014 [#permalink]
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11 Mar 2018, 12:51
Hi Bunuel
I solved this way. For denominator I figured out total cases = 4*4=16
For numerator I applied combinatorics: No. of ways digit 2 can be selected= 4C1 No. of ways digit other than 2 can be selected= 3C1 So total ways = 4C1*3C1
P= (4C1*3C1)/16= 3/4.
Please correct my mistake if possible, I suspect I am not considering the order, but how can I incorporate that here? Thanks so much.










