What is the probability of choosing 2 distinct positive

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Mon Jul 29, 2019 5:44 pm

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What is the probability of choosing 2 distinct positive integers from the set of first 10 positive integers such that their product and sum are both even?

A. 1/5
B. 2/9
C. 2/5
D. 4/9
E. 1/2

OA B

Source: e-GMAT

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Source: Beat The GMAT — Problem Solving | Mon Jul 29, 2019 5:44 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Mon Jul 29, 2019 9:48 pm

BTGmoderatorDC wrote:What is the probability of choosing 2 distinct positive integers from the set of first 10 positive integers such that their product and sum are both even?

A. 1/5
B. 2/9
C. 2/5
D. 4/9
E. 1/2

OA B

Source: e-GMAT

Given: There is a list of first 10 positive integers: 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

Note that the product of two integers is even if at least one between them is even. For example 2 and 3. Product of 2 and 3 = 2*3 = 6, even.

However, the sum of two integers is even if either both are even or both are odd. For example, 2 and 4. Sum of 2 and 4 = 2 + 4 = 6, even. Another example, 1 and 3. Sum of 1 and 3 = 1 + 3 = 4, even.

Thus, to maintain both the conditions that the sum, as well as the product, is even, we must have both the integers even.

# of ways of any choosing 2 even integers from {1, 2, 3, ..., 10} = 5C2 = 5.4/1.2 = 10; note that there 5 even integers

# of ways of any choosing 2 integers from {1, 2, 3, ..., 10} = 10C2 = 10.9/1.2 = 45;

Thus, the required probability = 10/45 = 2/9.

The correct answer: B

Hope this helps!

-Jay
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swerve: Legendary Member; Posts: 2545; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Tue Jul 30, 2019 10:38 am

First 10 positive integers $= \{1,2,3,4,5,6,7,8,9,10\}$

Ways to choose 2 positive integers $= 10C_2 = 45$.

The only group of even numbers can give sum and product to be even.

Possible groups are, $(2,4) (2,6) (2,8) (2,10) (4,6) (4,8) (4,10) (6,8) (6,10) (8,10) = 10$ groups.

Probability $= \frac{10}{45} = \frac{2}{9}$

__B__ is the correct answer.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

Re: What is the probability of choosing 2 distinct positive

by Brent@GMATPrepNow » Sat Oct 30, 2021 7:47 am

BTGmoderatorDC wrote: ↑
Mon Jul 29, 2019 5:44 pm
What is the probability of choosing 2 distinct positive integers from the set of first 10 positive integers such that their product and sum are both even?

A. 1/5
B. 2/9
C. 2/5
D. 4/9
E. 1/2

OA B

Source: e-GMAT

If two integers have an even product AND an even sum, then the two integers must both be even
So, the question is really asking us to find P(the 2 selected integers are both even).

The set of first 10 positive integers consists of 5 odd integers and 5 even integers.

So, we get: P(the 2 selected integers are both even) = P(the 1st integer is even AND the 2nd integer is even)
= P(the 1st integer is even) x P(the 2nd integer is even)
= 5/10 x 4/9
= 2/9

Answer: B

Brent Hanneson - Creator of GMATPrepNow.com

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