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

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

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%

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

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).

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%

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%

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.

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%

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.

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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%

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

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

What is wrong with my approach?

1-P(one even, one odd) = 1 - (5C1 * 5C1)/10C2

Hi ghnlrug,

The prompt tells us that we choose 2 balls "WITH REPLACEMENT" (meaning that after we choose the first ball, we put it back - and can potentially choose it again). The calculation 10c2 means that we can't choose the same ball twice - and that does not match up with what the prompts tells us.

GMAT assassins aren't born, they're made,
Rich

Thank you very much Rich.

So can I never use combinations whenever the question says "WITH REPLACEMENT"? Is that a good heuristic to remember?

Hi ghnlrug,

When a question involves 'replacement', I would not go so far as to say that I would "never" use the Combination Formula (since you might come across a quirky version of that type of question that could be approached with the Combination Formula) - but I would suggest suggest that you think about other ways to approach the calculation (and that usually means doing basic multiplication and 'mapping out' the possibilities).

GMAT assassins aren't born, they're made,
Rich