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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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New post 16 Feb 2015, 07:00
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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, 07:15
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
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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
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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.


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
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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, 07:56
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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
3
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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New post 17 Feb 2015, 19:18
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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
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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.


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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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
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]

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New post 14 Dec 2017, 05:47
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.


30-sec approach:

No matter what ball you pick for the first one, you'll have 1/2 chances to make the sum even with the second ball and the same 1/2 chances to make the sum odd.

Answer: 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 14 Sep 2018, 01:35
ok my take
urn contains 10 balls of which 5 are even and 5 are odd
P = number of choices with the sum even / all the possible choices of picking 2 numbers
here we have replacement and order matters ====> number or all possible choices are 10x10
for a sum of 2 numbers to be even it can be either the sum of 2 even integers or the sum of 2 odd integers :
number of choices = 5x5 + 5x5 = 50 choices (remember we have 5 even and 5 odd integers in the urn)
P = 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 27 Nov 2018, 19:26
1
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%

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.
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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 27 Nov 2018, 22:39
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

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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 27 Nov 2018, 23:16
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.


First, Total number of outcomes by drawing 2 balls with replacement = 10 X 10 = 100

for even sum outcomes , either both should be even or both should be odd.
For both even we get 5C1 * 5C1 = 25 outcomes , for odd 5C1 * 5C1 = 25 outcomes , so there are 50 possible outcomes

So , the 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 24 Mar 2019, 18:25
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 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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Re: An urn contains 10 balls, numbered from 1 to 10. If 2 balls are select   [#permalink] 24 Mar 2019, 18:25
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