An urn contains 10 balls, numbered from 1 to 10. If 2 balls

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

An urn contains 10 balls, numbered from 1 to 10. If 2 balls

by BTGmoderatorLU » Tue Nov 27, 2018 3:03 pm

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Source: Veritas Prep

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

The OA is C

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Source: Beat The GMAT — Problem Solving | Tue Nov 27, 2018 3:03 pm

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pm; Thanked: 59 times; Followed by:33 members

by fskilnik@GMATH » Tue Nov 27, 2018 6:16 pm

BTGmoderatorLU wrote:Source: Veritas Prep

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

There are exactly 4 equiprobable events:

E1 = first ball odd , second ball odd
E2 = first ball odd, second ball even
E3 = first ball even, second ball odd
E4 = first ball even, second ball even

Exactly two of the 4 events mentioned above are favorable (E1 and E4), hence the correct answer is 50% (C).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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by [email protected] » Tue Nov 27, 2018 9:18 pm

Hi All,

We're told that an urn contains 10 balls, numbered from 1 to 10 and we're told that 2 balls are selected at random WITH replacement. We're asked for the probability that the SUM of the 2 numbers on the balls will be EVEN. This question can be solved in a couple of different ways, including as a straight-probability question.

To start, there are two ways to get a sum that is EVEN:

(Odd on the first) and (Odd on the second)
(Even on the first) and (Even on the second)

Since there are an equal number of Odd and Even numbers AND we are replacing the first ball before we pull the second, the probability of pulling one Odd or one Even is the same each time: 1/2

(Odd) and (Odd) = (1/2)(1/2) = 1/4
(Even) and (Even) = (1/2)(1/2) = 1/4
Total probability of ending with a sum that is Even on two balls is 1/4 + 1/4 = 2/4 = 1/2

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Iakgmat: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Thu Dec 06, 2018 7:38 am

by Iakgmat » Fri Dec 07, 2018 7:36 am

There are 10 balls; 5 of which are even and 5 of which are odd.

Probability of drawing 1 even ball is 50%
Probability of drawing 1 odd ball is 50%

Here are the possible combos:
1. Draw even; Draw Even ==> Adds to Even
2. Draw odd; draw odd ==> adds to even

For (1): 50% * 50%
For (2): 50% * 50%

Then add them up, 25% + 25% = 50%

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Mar 24, 2019 5:09 pm

BTGmoderatorLU wrote:Source: Veritas Prep

An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even?

A. 25%
B. 37.5%
C. 50%
D. 62.5%
E. 75%

The OA is C

The sum of the 2 numbers will be even if the 2 numbers chosen are both even or both odd. The probability of choosing 2 even numbers is 1/2 x 1/2 = 1/4. Likewise, the probability of choosing 2 odd numbers is 1/2 x 1/2 = 1/4. Therefore, the probability of choosing 2 numbers whose sum is even is 1/4 + 1/4 = 1/2.

Answer: C

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Founder and CEO
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