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

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

by BTGmoderatorDC » Sat Jul 20, 2019 4:17 pm

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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
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by Darkknightreturning » Sat Jul 20, 2019 9:18 pm
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%.
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by Darkknightreturning » Sat Jul 20, 2019 9:23 pm
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
Ok. since the question includes the term "with replacement". Thus, once one ball has been taken out, it also gets replaced.
So the number of ways in which odd numbered balls are selected = 5x5 (since the ball gets replaced)
the number of ways in which even numbered balls are selected = 5x5 (since the ball gets replaced)

Total ways of selecting 2 balls out of 10 balls with replacement = 10x10
Thus, the probability of getting balls with even sum = (25+25)/100= 50%
Thus, C is the correct answer.
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by Brent@GMATPrepNow » Sun Jul 21, 2019 4:53 am
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

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Brent
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by Scott@TargetTestPrep » Tue Jul 30, 2019 8:38 am
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

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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