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# A box contains 100 balls, numbered from 1 to 100. If three b

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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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16 Aug 2014, 03:26
E = Even, O = Odd
EEO = 1st ball is even, 2nd ball is even, 3rd ball is odd
Prob(EEO) + Prob(EOE) + Prob(OEE) + Prob(OOO) = 4*Prob(EEO) = 4 * Prob(E)*Prob(E)*Prob(O) = 4*(1/2)^3 = 4/8 = 1/2
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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16 Nov 2015, 10:41
I reasoned this out the following way, is it correct? Regardless of the order we have 4 total scenarios (combinations) of even-odd that can impact our game:

1. all even
2. all odd (works for us)
3. 2 even 1 odd (works for us)
4. 2 odd 1 even.

The order does not matter because in any order (whether it is 2-2-1 or 2-1-2) we will get an odd number in 2 combinations out 4 possible.

Hence 1/2
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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24 Nov 2015, 08:41
*laughing**
there is a quick backdoor here.
whatever the sum is, the probability of the number being odd is 50/100.
So what are we solving here?
the answer jumps right at you.
50/100 = 1/2.
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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13 Mar 2016, 15:21
I unknowingly stumbled into this answer. On probability questions, I always look to solve backwards since the GMAT often offers that shortcut.

I said what is the probability that the sum will be even.

E+E+E or O+O+E

Then I realized that to be odd, it will be

O+O+O or E+E+O

Hence they're both 1/2
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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31 Mar 2016, 00:53
Attached is a visual that should help. This is a different method from the one in the book, but I find it to be more logical.
Attachments

Screen Shot 2016-03-31 at 12.52.48 AM.png [ 113.15 KiB | Viewed 3226 times ]

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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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15 Jun 2016, 21:18
Given: 100 balls. 3 balls are selected with replacement from the box.
Required: Sum of the balls has to be odd.

The sum will be odd if:

3balls odd
2 even 1 odd

Total cases: 3odd, 1even 2 odd, 2 even 1 odd, 3 odd = 4

Probability that the sum will be odd = 2/4 = 1/2

Correct Option: C
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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17 Jun 2016, 07:46
Original question is:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

The sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls (Odd+Odd+Odd=Odd) or two even numbered balls and one odd numbered ball (Even+Even+Odd=Odd);

P(OOO)=(1/2)^3;
P(EEO)=3*(1/2)^2*1/2=3/8 (you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways: EEO, EOE, or OEE);

So finally P=1/8+3/8=1/2.

Hi,

Why would you need to multiply by 3 even if the scenario occurs in 3 ways. What is the use of doing it? Adding any numbers in OEE or EEO or EOE format would only produce an odd number as result. Am i missing something here?

Further i did it in this way. Can you confirm if this is fine?

There are only 4 outcomes when you have to choose Evens or Odds for 3 numbers.

1. OOO
2. EEE
3. OEE
4. EOO

Of the above, the sum of 2 results in odd numbers. Hence favourable ways/Total ways = 2/4 = 1/2

Am i right ?

Thanks

Alternately you can notice that since there are equal chances to get even or odd sum after two selections (for even sum it's EE or OO and for odd sum it's EO or OE) then there is 1/2 chances the third ball to make the sum odd.[/quote]
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A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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20 Jun 2016, 03:42
There is a much simper way to approach this problem. You will get the answer in 5 seconds

In this question you do not have to worry about making cases (E+E+O) or (E+O+O) or (O+E+O) or any such combination.
When you have done all the fancy juggling, replacement, non replacement, this that blah blah.. what will happen
At the end of the day you will either have an odd or even.
Thus your total outcome can be 2 and your favoured outcome is 1
The sample space here is 2 (either odd or even) and the favoured outcome is 1 (odd)

$$Probability = \frac{Favorable Outcome}{Sample Space}$$

$$Probality (odd)= \frac{Odd}{Odd+Even}$$

$$Probality (odd)= \frac{1}{2}$$

Consider this :- Getting an odd or an even number is a mutually exclusive event or binomial event.

In terms of divisibility by 2 how many kind of number exist
(a) Divisible by 2 => Even
(b) Not divisible by 2=> Odd

No matter how many numbers in what way you add, the end result will always be either odd or even.

Adding only even numbers or adding only odd numbers or adding both odd and even numbers does not change the end result which is either odd or even
Adding 10 numbers or adding 19487500 numbers or adding 10000000000000 numbers or adding 123467856453890182800 numbers will not change the sample space i.e EVEN or ODD i.e 2
7+3=10 EVEN
5+3+1=9 ODD
4+6+8+10=28 EVEN
3+7+9+1+5=25 ODD
1+2+3+4+5+6=21 ODD
1+2+3+4+5+6+7=28 EVEN

Hence probability = $$\frac{Favorable outcome}{Total number of outcome}$$
Probability (odd)= odd/odd+even =$$\frac{1}{2}$$

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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01 Jun 2017, 00:47
Bunuel wrote:
mbafall2011 wrote:

Since replacement is involved, i would think the order of the EEO ball being picked does not matter.

Thus P(E)&P(E)&P(O) should be 1/8

1/8+1/8 = 1/4.

OA is 1/2.

It doesn't matter for order whether it's with or without replacement case. EEO, EOE, and OEE are 3 different scenarios and each has the probability of 1/8, so the probability of two even numbered balls and one odd numbered ball is 3*1/8.

Total: 1/4+3/8=1/2.

1. A ball can be odd or even numbered, for total number of 8 possibilities for 3 selections
2. No ball odd and 3 balls even, sum being even and occurring once
3. 1 ball being odd and 2 balls even, sum being odd and occurring thrice
4. 2 balls being odd and 1 even, sum being even and occurring thrice
5. All balls being odd, sum being odd occurring once
6. 4 times the sum is even and 4 times it is odd for a probability of 3/6=1/2.
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A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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10 Jun 2017, 07:44
In general there are the following cases:

1) All three balls are odd: Result is odd (e.g. 3+3+3=9=odd)
2) One ball is odd and the other two are even: Result is odd (e.g. 3+2+2=7= odd)

3) One ball is even and the other two are odd: Result is even (e.g. 2+3+3=8=even)
4) All three balls are even: Result is even (e.g. 2+2+2=6=even)

Therefore the probability that the result will be odd is 2/4 = 1/2. Answer Choice (C)

Note: The number of ways in which the second scenario can occur is identical with the number of ways in which the third scenario can occur. The same applies for scenario 1) and 4). Therefore, it is negligible.
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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14 Dec 2017, 07:00
As the no. of odd integers= no. of even integers= 50.
So by symmetry, Probability of getting sum as Even = Probability of getting sum as Odd = P (suppose)
Also, Probability of getting sum as Even + Probability of getting sum as Odd = 1 ( as they mutually exclusive and exhaustive events)
thus P+P = 1
or, 2P =1
or, P=1/2

PS:
Note 1: This is applicable to both replacement and without replacement case.
Note 2: This is not applicable to the case when the number of odd integers is not equal to the number of even integers as then the condition will not be symmetric w.r.t odd and even integers.
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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28 Dec 2017, 06:08
Bunuel wrote:
Bunuel wrote:
Again order does matter. P(odd sum)=P(EEO)+P(EOE)+P(OEE)+P(OOO)=1/8+1/8+1/8+1/8=1/2.

Excuse me, but I didn't understand why order does matter? At the end we are looking for the sum of the selected balls and not for the order of the selection, so whether it is 2+2+1 or 1+2+2 or 2+2+1 they are all the same! They all equal 4 which is one possible outcome and not 3

Consider below two scenarios:
First=Even, Second=Even, Third=Odd;
First=Even, Second=Odd, Third=Even;

Are these scenarios the same? No. That's why the order matters.

Hi

Could you please give me the link of this type of questions so that I can learn better. I need to practice this type of questions in which probability of two will be different.
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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11 Jan 2018, 16:47
Hi All,

These types of questions can be solved in a couple of different ways, depending on how you like to organize your information and how you "see" the question.

Since half the numbered balls are 'odd' and half are 'even', and we're replacing each ball after selecting it, on any given selection we have a 50/50 chance of getting an odd or even. This means that there are....

(2)(2)(2) = 8 possible arrangements when selecting 3 balls from the box:

EEE
EEO
EOE
OEE

OOO
OOE
OEO
EOO

Since we have the same number of Odds and Evens, each of these options has the same 1/8 probability of happening. Now you just have to find the ones that give us the ODD SUM that we're asked for. They are:

EEO
EOE
OEE
OOO

Four of the eight options. 4/8 = 1/2

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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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10 Sep 2018, 05:20
Bunuel wrote:
SmitKhurana wrote:
Hello there GMAT enthusiasts!

Surely this finds everyone in great guns towards achieving a perfect GMAT Score, in between came across this very peculiar and relatively difficult question for resolution :

Q. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ?

Welcome to GMAT Club!

Provide answer choices for PS questions.

Original question is:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

The sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls (Odd+Odd+Odd=Odd) or two even numbered balls and one odd numbered ball (Even+Even+Odd=Odd);

P(OOO)=(1/2)^3;
P(EEO)=3*(1/2)^2*1/2=3/8 (you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways: EEO, EOE, or OEE);

So finally P=1/8+3/8=1/2.

Alternately you can notice that since there are equal chances to get even or odd sum after two selections (for even sum it's EE or OO and for odd sum it's EO or OE) then there is 1/2 chances the third ball to make the sum odd.

How will you solve this problem using combinatorial way?
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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04 Nov 2018, 12:39
nitinkarnwal wrote:
Bunuel wrote:
SmitKhurana wrote:
Hello there GMAT enthusiasts!

Surely this finds everyone in great guns towards achieving a perfect GMAT Score, in between came across this very peculiar and relatively difficult question for resolution :

Q. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ?

Welcome to GMAT Club!

Provide answer choices for PS questions.

Original question is:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

The sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls (Odd+Odd+Odd=Odd) or two even numbered balls and one odd numbered ball (Even+Even+Odd=Odd);

P(OOO)=(1/2)^3;
P(EEO)=3*(1/2)^2*1/2=3/8 (you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways: EEO, EOE, or OEE);

So finally P=1/8+3/8=1/2.

Alternately you can notice that since there are equal chances to get even or odd sum after two selections (for even sum it's EE or OO and for odd sum it's EO or OE) then there is 1/2 chances the third ball to make the sum odd.

How will you solve this problem using combinatorial way?

Hi Bunuel,

I have the same doubt- How can we solve this problem in combinatorial approach?
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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24 Feb 2019, 09:25
Combinatorics solution:

Amount of ways I can pick, then order, 3 odds or even: 2^3=8
Amount of ways I can pick 1 odd and 2 even: OEE can be ordered in 3!/2!=3
Amount of ways I can pick 3 odd: OOO can be ordered in 3!/3!=1

Overall probability: 4/8=1/2.
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A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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21 May 2019, 03:34
The easiest way to approach is-

The possible ways for the sum to be odd are

O O O

or

E E O

Now, the point to notice is- The 3rd draw i.e the 3rd ball will be the determinant whether the sum will be odd or even ( you can try to replace the last position holder and see the result changes) and since there is an equal probability of odd/even to occur i.e. 1/2 so the answer will be 1/2.

expanding a little more on my explanation, quickly draw 3 dashes ( as above) and fill out the first two place holders with an entity of your choice (odd/even), now the 3rd place holder you'll fill will make all the difference i.e. it will either change the result to odd/even and the probability of both events is 1/2.
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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09 Jul 2019, 11:04
SmitKhurana wrote:
A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

total even = 50 and odd ; 50 so P even ; 50/100 ; 1/2 and same for odd 1/2
we need to find three balls drawn at random and sum = odd ; only possible when
case 1; e+e+o ; can be arranged in 3!/2! ; 3 ways ; 3*(1/2)^2*1/2 ; 3/8 ways
and case 2; o+o+o ; 1/2^3 ; 1/8
total 1/8+3/8 ; 4/8 ; 1/2
IMO C
Re: A box contains 100 balls, numbered from 1 to 100. If three b   [#permalink] 09 Jul 2019, 11:04

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