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

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An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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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%

Kudos for a correct solution.
[Reveal] Spoiler: OA

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An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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Bunuel 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%

Kudos for a correct solution.


Probability = Favorable outcome / Total Outcome

Total Outcomes = 10 x 10 = 100 (there will be 10 possibilities each time 'With replacement')

Favorable Outcome will be obtained when either both numbers are even or both numbers selected are odd. There are 5 even and 5 odd numbered balls for each selection of ball

1) Both Even: Therefore total outcomes = 5 x 5 = 25
2) Both Odd: Therefore total outcomes = 5 x 5 = 25

Total Favorable Outcomes = 25+25 = 50

Probability = 50/100 = 1/2 = 50%

Answer: Option
[Reveal] Spoiler:
C

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An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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New post 16 Feb 2015, 07:24
Bunuel 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%

Kudos for a correct solution.


Alternate Method:

Probability = Favorable outcome / Total Outcome

Favorable Outcome will be obtained when EITHER both numbers are even OR both numbers selected are odd.

EVEN + EVEN = EVEN
OR
ODD + ODD = EVEN

Now,

Case 1: BOTH EVEN
Step 1) Probability of First ball being Even numbered = 5/10 = 1/2 (as there are 10 balls to pick from which have 5 even numbered and 5 odd numbered)
Step 2) Probability of Second ball being Even numbered = 5/10 = 1/2 (as there are 10 balls to pick from due to the condition 'WITH REPLACEMENT' which have 5 even numbered and 5 odd numbered)

Hence Probability of the sum being even as per Case-1 = (1/2) x (1/2) = 1/4


Case 2: BOTH ODD
Step 1) Probability of First ball being Odd numbered = 5/10 = 1/2 (as there are 10 balls to pick from which have 5 even numbered and 5 odd numbered)
Step 2) Probability of Second ball being Odd numbered = 5/10 = 1/2 (as there are 10 balls to pick from due to the condition 'WITH REPLACEMENT' which have 5 even numbered and 5 odd numbered)

Hence, Probability of the sum being even as per case-2 = (1/2) x (1/2) = 1/4


Total Probability = (1/4) + (1/4) = 1/2 = 50%

Answer: Option
[Reveal] Spoiler:
C

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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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actually in this particular question we donot even require to calculate..
as we have equal number of odd and even number, the sum will be half the time even and half the time odd..
so ans 50%... C

also the numbers can be picked in four ways..
1)odd,odd
2)even,even
3)odd,even
4)even,odd
out of these 1 and 2 will give even number.. 3 and 4 will give odd number
so 2/4=50%..
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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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New post 17 Feb 2015, 17:54
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this is a straightforward C!
5 even 5 odds
So the first ball will have = 5/10 to be odd * 5/10 to have the second one odd
+
5/10 to be even * 5/10 to have the second one even
=
50%
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An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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Bunuel 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%

Kudos for a correct solution.



The Selection of balls:

Either both are even or odd

so in each case there is equal no - 5

Probability of drawing Even no one after one: 5/10 *5/10

Probability of drawing ODD no one after one: 5/10 *5/10

Either of them is possible, so Probility so that sum is either odd or even : ( 5/10 *5/10) + ( 5/10 *5/10) = 1/2 = 50%

Ans C

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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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New post 22 Feb 2015, 11:27
Bunuel 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%

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Correct Answer: C

Since there are 5 even numbered balls and 5 odd numbered balls, the probability of selecting either an even or an odd is 5/10= 1/2. In order for the sum of the numbers on the two balls to be even, either both numbers must be even or both numbers must be odd; one even and one odd number would add to an odd number. The probability of selecting two even numbers is:

1/2 * 1/2= 1/4

The probability of selecting two odd numbers is also:

1/2 * 1/2= 1/4

Adding both probabilities gives 1/4+1/4= 1/2 , or 50%, as the probability that the sum of the numbers will be even. Thus, the correct answer is (C).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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New post 22 Sep 2017, 00:10
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Bunuel 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%

Kudos for a correct solution.



Probability = Favorable outcome / Total Outcome

Total Outcomes = 10 x 10 = 100 (there will be 10 possibilities each time 'With replacement')

Favorable Outcome will be obtained when either both numbers are even or both numbers selected are odd. There are 5 even and 5 odd numbered balls for each selection of ball

1) Both Even: Therefore total outcomes = 5 x 5 = 25
2) Both Odd: Therefore total outcomes = 5 x 5 = 25

Total Favorable Outcomes = 25+25 = 50

Probability = 50/100 = 1/2 = 50%

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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select [#permalink]

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New post 22 Sep 2017, 08:58
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?

Number of even values = 5
Number of odd values = 5

Total number of possibilities = 10*10 (chances of picking 10 at each go)

Even plus even and odd plus odd returns an even value, hence, total number of favorable outcomes = 5*5+5*5/ 100 = 1/2 = 50%

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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select   [#permalink] 22 Sep 2017, 08:58
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