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M25-17

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New post 16 Sep 2014, 00:23
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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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New post 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 \(\frac{4}{9}\).


Answer: C
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Re: M25-17  [#permalink]

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New post 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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Kudos if this was helpful, please (:
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Re: M25-17  [#permalink]

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New post 12 Apr 2015, 13:13
How did 9 come?
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New post 13 Apr 2015, 03:00
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M25-17  [#permalink]

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New post 20 Jul 2016, 14:45
Easiest and fastest, imo

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Re: M25-17  [#permalink]

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New post 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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New post 20 Jun 2018, 11:49
This was a pretty simple one.

(1/3)*(1/3)*4 = 4/9
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Re: M25-17  [#permalink]

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New post 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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New post 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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New post 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 \(\frac{4}{9}\).


Answer: C


Why isn total outcomes = 6c2 correct?
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New post 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 \(\frac{4}{9}\).


Answer: C


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

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