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Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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Difficulty:   45% (medium)

Question Stats: 67% (02:03) correct 33% (02:58) wrong based on 84 sessions

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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. $$\frac{5}{8}$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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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?".

$$P(\text{the sum is odd}) = (P(\text{the first number is 2}) * P(\text{the second number is not 2})) +$$ $$+ (P(\text{the first number is not 2}) * P(\text{the second number is 2})) =$$ $$= \frac{1}{4}*\frac{3}{4} + \frac{3}{4}*\frac{1}{4} = \frac{6}{16} = \frac{3}{8}$$.

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Intern  Joined: 30 Jun 2012
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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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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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Manager  B
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 230
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32 GMAT 2: 680 Q47 V35 ### Show Tags

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 sub-600 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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Intern  Joined: 03 Jul 2016
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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?
Manager  S
Joined: 17 Aug 2015
Posts: 94
GMAT 1: 650 Q49 V29 ### Show Tags

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
Intern  Joined: 02 Sep 2016
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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.
Intern  Joined: 25 Jul 2016
Posts: 8
GMAT 1: 740 Q50 V40 ### Show Tags

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*$$\frac{1}{4}$$*$$\frac{1}{4}$$=$$\frac{3}{8}$$
Current Student S
Joined: 01 Dec 2016
Posts: 107
Concentration: Finance, Entrepreneurship
GMAT 1: 650 Q47 V34 WE: Investment Banking (Investment Banking)

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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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Intern  B
Joined: 28 Apr 2016
Posts: 19

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The question says "not necessarily different". Can we safely assume that the first number will be pot in the box again?
Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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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.
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Intern  B
Joined: 28 Apr 2016
Posts: 19

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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dyg wrote:
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.

If it is in this context then yes.
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Intern  B
Joined: 10 May 2017
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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
Intern  B
Joined: 29 Jul 2016
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I think this is a poor-quality 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.

Math Expert V
Joined: 02 Sep 2009
Posts: 53792

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Sushilait84 wrote:
I think this is a poor-quality 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.

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/m10-183859.html#p1452001
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Manager  G
Joined: 12 Jun 2016
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Hello Bunuel

I think this is a high-quality 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/m20-184246.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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Intern  B
Joined: 14 May 2017
Posts: 47

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P (EO) = 1/4 *3/4* 2! = 3/8

If you have any confusion with this method let me know.
Intern  B
Joined: 08 Aug 2017
Posts: 31

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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. Re: M10-14   [#permalink] 11 Mar 2018, 12:51

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