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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An urn contains 10 balls, numbered from 1 to 10. If 2 balls
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BTGmoderatorDC
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Darkknightreturning
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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%.
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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Darkknightreturning
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Ok. since the question includes the term "with replacement". Thus, once one ball has been taken out, it also gets replaced.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
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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There are 2 possible ways to get an EVEN sum: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
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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Scott@TargetTestPrep
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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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