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

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BTGmoderatorLU: Moderator; Posts: 2598; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Fri Jan 12, 2018 2:27 pm

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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Source: Beat The GMAT — Problem Solving | Fri Jan 12, 2018 2:27 pm

elias.latour.apex: Master | Next Rank: 500 Posts; Posts: 228; Joined: Thu Apr 20, 2017 1:02 amFollowed by: 3 members; Location: Global; Thanked: 32 times; GMAT Score:770

by elias.latour.apex » Sat Jan 13, 2018 6:06 am

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.

Elias Latour
Verbal Specialist @ ApexGMAT
blog.apexgmat.com
+1 (646) 736-7622

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8198; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Aug 11, 2019 6:53 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 a and b both odd or both 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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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