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16 Sep 2014, 01:23
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Two sets are defined as follows:
\(X = \{4, 5, 8\}\)
\(Y = \{-1, 1, 2\}\)
If a number is randomly selected from set \(X\) and another number is randomly selected from set \(Y\), what is the probability that their sum will be an even integer?
A. \(\frac{2}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{4}{9}\)
D. \(\frac{5}{9}\)
E. \(\frac{2}{3}\)
\(X = \{4, 5, 8\}\)
\(Y = \{-1, 1, 2\}\)
If a number is randomly selected from set \(X\) and another number is randomly selected from set \(Y\), what is the probability that their sum will be an even integer?
A. \(\frac{2}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{4}{9}\)
D. \(\frac{5}{9}\)
E. \(\frac{2}{3}\)
16 Sep 2014, 01:23
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Official Solution:
Two sets are defined as follows:
\(X = \{4, 5, 8\}\)
\(Y = \{-1, 1, 2\}\)
If a number is randomly selected from set \(X\) and another number is randomly selected from set \(Y\), what is the probability that their sum will be an even integer?
A. \(\frac{2}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{4}{9}\)
D. \(\frac{5}{9}\)
E. \(\frac{2}{3}\)
The total number of pairs possible is 3*3 = 9. To get an even sum, either both selected numbers must be even or both selected numbers must be odd. Hence, the four possible pairs giving an even sum are: (4, and 2); (5 and -1); (5 and 1); and (8 and 2). Therefore, the probability that the sum is even is \(\frac{4}{9}\).
Answer: C
Two sets are defined as follows:
\(X = \{4, 5, 8\}\)
\(Y = \{-1, 1, 2\}\)
If a number is randomly selected from set \(X\) and another number is randomly selected from set \(Y\), what is the probability that their sum will be an even integer?
A. \(\frac{2}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{4}{9}\)
D. \(\frac{5}{9}\)
E. \(\frac{2}{3}\)
The total number of pairs possible is 3*3 = 9. To get an even sum, either both selected numbers must be even or both selected numbers must be odd. Hence, the four possible pairs giving an even sum are: (4, and 2); (5 and -1); (5 and 1); and (8 and 2). Therefore, the probability that the sum is even is \(\frac{4}{9}\).
Answer: C
General Discussion
08 Nov 2014, 15:10
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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
------------------------------------------------------------------------------------------------------------
Kudos if this was helpful, please (:
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
------------------------------------------------------------------------------------------------------------
Kudos if this was helpful, please (:
