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# M25-17

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Math Expert
Joined: 02 Sep 2009
Posts: 65096

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16 Sep 2014, 00:23
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Difficulty:

5% (low)

Question Stats:

82% (01:09) correct 18% (01:27) wrong based on 284 sessions

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Two sets are defined as follows:
$$X = \{4, 5, 8\}$$
$$Y = \{-1, 1, 2\}$$

If a number is taken from set $$X$$ at random and another number is taken from set $$Y$$ at random, what is the probability that the sum of these numbers will be an even integer?

A. $$\frac{2}{9}$$
B. $$\frac{1}{3}$$
C. $$\frac{4}{9}$$
D. $$\frac{5}{9}$$
E. $$\frac{2}{3}$$

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Joined: 02 Sep 2009
Posts: 65096

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16 Sep 2014, 00:23
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Official Solution:

Two sets are defined as follows:
$$X = \{4, 5, 8\}$$
$$Y = \{-1, 1, 2\}$$

If a number is taken from set $$X$$ at random and another number is taken from set $$Y$$ at random, what is the probability that the sum of these numbers will be an even integer?

A. $$\frac{2}{9}$$
B. $$\frac{1}{3}$$
C. $$\frac{4}{9}$$
D. $$\frac{5}{9}$$
E. $$\frac{2}{3}$$

There are four favorable outcomes (4+2, 5+1, 5-1, 8+2) out of 9 possible. Therefore, the probability is $$\frac{4}{9}$$.

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Joined: 08 May 2014
Posts: 10
Schools: HBS '19 (D)
GRE 1: Q164 V165

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08 Nov 2014, 14:10
5
I did it this way:

You need EVEN+EVEN or ODD+ODD to have an even result.

X = 2 even numbers (4,8) and 1 odd number (5)
Y = 2 odd numbers (1,-1) and 1 even number (2)

You know that you want consecutive events (draw one number from X and then one number of Y)

1. Probability of EVEN+EVEN = (2/3)*(1/3) - you have 2 out of 3 in set X that are even (4,8) and 1 out of 3 in set Y that are even (2). Since the events happen "at the same time" or one after the other, you multiply them (you need one even from set X AND one from set Y)

2. Probability of ODD+ODD = (1/3)*(2/3) - you have 1 out of 3 that is odd in set X and 2 that are odd in set Y that are odd

Sum both (you need EVEN+EVEN OR ODD+ODD, so you sum them)

(2/9)+(2/9) = 4/9

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Intern
Joined: 12 Jan 2015
Posts: 14
Concentration: Entrepreneurship, Human Resources
GMAT Date: 06-27-2015

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12 Apr 2015, 13:13
How did 9 come?
Math Expert
Joined: 02 Sep 2009
Posts: 65096

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13 Apr 2015, 03:00
SonaliT wrote:
How did 9 come?

There are 3 numbers in each set, so there are 3*3=9 different pairs.
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20 Jul 2016, 14:45
Easiest and fastest, imo

http://imgur.com/tmIwKSH
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Joined: 08 Jun 2015
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GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
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20 Jun 2018, 07:06
The required probability is = #of possible cases/total # of outcomes. #of possible cases is 4. Total # of outcomes is 3*3 = 9. Therefore the answer is 4/9 or option C.
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20 Jun 2018, 11:49
This was a pretty simple one.

(1/3)*(1/3)*4 = 4/9
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21 Jun 2018, 05:36
Probability of selecting 4 from set X = P(4) = 1/3

The probability of selecting 2 from set Y = P(2) = 1/3

So the required probability can be written as

P(4)*P(2) + P (5) * P (1 or -1) + P (8) * P (2 )

= (1/3)*(1/3) + (1/3)*(2/3) + (1/3)*(1/3)

= 4/9
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Joined: 04 Jul 2017
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21 Jun 2018, 07:54
SonaliT wrote:
How did 9 come?

9 = 3 numbers of set X times 3 numbers of set Y which represents to total number of selecting a number of set X and a number from set Y
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Joined: 08 Aug 2019
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02 Jun 2020, 22:32
Bunuel wrote:
Official Solution:

Two sets are defined as follows:
$$X = \{4, 5, 8\}$$
$$Y = \{-1, 1, 2\}$$

If a number is taken from set $$X$$ at random and another number is taken from set $$Y$$ at random, what is the probability that the sum of these numbers will be an even integer?

A. $$\frac{2}{9}$$
B. $$\frac{1}{3}$$
C. $$\frac{4}{9}$$
D. $$\frac{5}{9}$$
E. $$\frac{2}{3}$$

There are four favorable outcomes (4+2, 5+1, 5-1, 8+2) out of 9 possible. Therefore, the probability is $$\frac{4}{9}$$.

Why isn total outcomes = 6c2 correct?
Math Expert
Joined: 02 Sep 2009
Posts: 65096

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02 Jun 2020, 23:19
AbenGeorge wrote:
Bunuel wrote:
Official Solution:

Two sets are defined as follows:
$$X = \{4, 5, 8\}$$
$$Y = \{-1, 1, 2\}$$

If a number is taken from set $$X$$ at random and another number is taken from set $$Y$$ at random, what is the probability that the sum of these numbers will be an even integer?

A. $$\frac{2}{9}$$
B. $$\frac{1}{3}$$
C. $$\frac{4}{9}$$
D. $$\frac{5}{9}$$
E. $$\frac{2}{3}$$

There are four favorable outcomes (4+2, 5+1, 5-1, 8+2) out of 9 possible. Therefore, the probability is $$\frac{4}{9}$$.

Why isn total outcomes = 6c2 correct?

Because we are not choosing 2 out of 6. 6C2 will contain pairs which are not possible, for example, (4, 5), (4, 8), (5, 8), (-1, 1), ... We are choosing one number from $$\{4, 5, 8\}$$, 3 options, and one number from $$\{-1, 1, 2\}$$, also 3 options. Total = 3*3 = 9.
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Re: M25-17   [#permalink] 02 Jun 2020, 23:19

# M25-17

Moderators: chetan2u, Bunuel