If a is to be chosen at random from the integers between 1 to 5, inclusive, and b is to be chosen at random from the integers between 6 and 10, inclusive, what is the probability that a + b will be even?
(A) 6/25
(B) 2/5
(C) 12/25
(D) 3/5
(E) 16/25
The OA is C.
I'm really confused with this PS question. Experts, any suggestion? Thanks in advance.
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If a is to be chosen at random from the integers...
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BTGmoderatorLU
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elias.latour.apex
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For a+b to be even, either both a and b must be even or both a and b must be odd.
Since we will be choosing 5 possible "a" numbers and 5 possible "b" numbers, the total number of possibilities will be 25. The question is: How many of those will fit the above criteria?
We have 3 ways that "a" can be odd and 2 ways that "b" can be odd. So that is 6 possible ways (3x2).
We have 2 ways that "a" can be even and 3 ways that "b" can be even. So that is 6 possible ways.
So, all told, we have 12 possible successes out of 25 possible choices. So the probability will be 12/25.
Since we will be choosing 5 possible "a" numbers and 5 possible "b" numbers, the total number of possibilities will be 25. The question is: How many of those will fit the above criteria?
We have 3 ways that "a" can be odd and 2 ways that "b" can be odd. So that is 6 possible ways (3x2).
We have 2 ways that "a" can be even and 3 ways that "b" can be even. So that is 6 possible ways.
So, all told, we have 12 possible successes out of 25 possible choices. So the probability will be 12/25.
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Scott@TargetTestPrep
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In order for a + b to be even, we must have a and b both odd or both even.BTGmoderatorLU wrote:If a is to be chosen at random from the integers between 1 to 5, inclusive, and b is to be chosen at random from the integers between 6 and 10, inclusive, what is the probability that a + b will be even?
(A) 6/25
(B) 2/5
(C) 12/25
(D) 3/5
(E) 16/25
P(a and b are both odd) = 3/5 x 2/5 = 6/25
P(a and b are both even) = 2/5 x 3/5 = 6/25
So P(a + b = even) = 6/25 + 6/25 =12/25.
Answer: C
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