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If a is to be chosen at random from the integers between

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by BTGmoderatorLU » Fri Apr 13, 2018 2:01 pm

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Answers

A

B

C

D

E

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Difficulty—

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.

Please, can anyone help me to solve this PS question? Thanks in advance.
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by [email protected] » Fri Apr 13, 2018 4:55 pm
Hi All,

We're told that "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. We're asked for the probability that A+B will be EVEN. This question is based on some Number Properties and Probability math.

To start, there are two ways for the sum of two integers to be EVEN:
1) BOTH integers are EVEN. The probability of that occurring here is (2/5)(3/5) = 6/25
2) BOTH integers are ODD. The probability of that occurring here is (3/5)(2/5) = 6/25

Thus, the overall probability of either of those two outcomes happening is 6/25 + 6/25 = 12/25.

Final Answer: C

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by Scott@TargetTestPrep » Wed Apr 25, 2018 3:50 pm
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
In order for (a + b) to be even we must have either both a and b are odd or both a and b are even.

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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