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

Question

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%

OA C

Source: Veritas Prep

Darkknightreturning · Answer

For the sum to be even, the balls selected must be both odd or both even.
Both odd numbered balls will mean, balls with numbers 1,3,5,7,9 i.e. 5
Both even numbered balls will mean balls with numbers 2,4,6,8,10 i.e. 5

Any other combination will yield an odd sum.
For selecting 2 odd numbered balls, number of ways will be 5C2
For selecting 2 even numbered balls, number of ways will be 5C2
Total ways of selecting 2 balls out of 10 balls = 10C2

Thus, probability of getting balls with even sum = (5C2+5C2)/10C2 which works out to be 44.44%.

Darkknightreturning · Answer

BTGmoderatorDC wrote: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%

OA C

Source: Veritas Prep

There are 2 possible ways to get an EVEN sum:
1) The 1st and 2nd balls are both EVEN
2) The 1st and 2nd balls are both ODD

So, P(sum is even) = P(1st is even AND 2nd is even OR 1st is odd AND 2nd is odd)
= P(1st is even AND 2nd is even) + P(1st is odd AND 2nd is odd)
= [P(1st is even) x P(2nd is even)] + [P(1st is odd) x P(2nd is odd)]
= [1/2 x 1/2] + [1/2 x 1/2]
= 1/4 + 1/4
= 1/2
= 0.5
= 50%

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

BTGmoderatorDC wrote: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%

OA C

Source: Veritas Prep

There are 2 possible ways to get an EVEN sum:
1) The 1st and 2nd balls are both EVEN
2) The 1st and 2nd balls are both ODD

So, P(sum is even) = P(1st is even AND 2nd is even OR 1st is odd AND 2nd is odd)
= P(1st is even AND 2nd is even) + P(1st is odd AND 2nd is odd)
= [P(1st is even) x P(2nd is even)] + [P(1st is odd) x P(2nd is odd)]
= [1/2 x 1/2] + [1/2 x 1/2]
= 1/4 + 1/4
= 1/2
= 0.5
= 50%

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Scott@TargetTestPrep · Answer

BTGmoderatorDC wrote: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%

OA C

Source: Veritas Prep

There are 2 possible ways to get an EVEN sum:
1) The 1st and 2nd balls are both EVEN
2) The 1st and 2nd balls are both ODD

So, P(sum is even) = P(1st is even AND 2nd is even OR 1st is odd AND 2nd is odd)
= P(1st is even AND 2nd is even) + P(1st is odd AND 2nd is odd)
= [P(1st is even) x P(2nd is even)] + [P(1st is odd) x P(2nd is odd)]
= [1/2 x 1/2] + [1/2 x 1/2]
= 1/4 + 1/4
= 1/2
= 0.5
= 50%

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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