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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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New post 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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New post 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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New post 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.
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Screen Shot 2016-03-31 at 12.52.48 AM.png
Screen Shot 2016-03-31 at 12.52.48 AM.png [ 113.15 KiB | Viewed 1931 times ]


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

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

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New post 17 Jun 2016, 07:59
1
nishi999 wrote:
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]

Hi,

Your way is also correct .......

the difference in yours and the other method is that you are not considering ORDER and in other solution ORDER is considered....
Just remember you will get the answer here BOTH ways.. BUT whatever you start with , continue with it..

You did not consider order, so you should not consider till end..

If order was to be considered in your method..
1. OOO- Only 1 way
2. EEE - again ONLY one way
3. OEE - 3 ways - OEE, EOE, EEO
4. EOO - 3 ways - OOE, OEO, EOO

total 8 ways.. you are interested in serial number 1 and 4, so 1+3 = 4 ways..
Prob = 4/8 = 1/2 ways
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A box contains 100 balls, numbered from 1 to 100. If three b [#permalink]

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New post 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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New post 01 Jun 2017, 00:47
Bunuel wrote:
mbafall2011 wrote:
What is the official answer.

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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New post 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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New post 11 Dec 2017, 02:46
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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


METHOD 1:

CASE-1: All odd: (1/2)*(1/2)*(1/2)=1/8

CASE-2: Two even one odd: 3C2*(1/2)*(1/2)*(1/2)= 3/8

Total = (1/8)+(3/8)= (1/2)


METHOD 2:

Or

There are as many even numbers as odd numbers from 1-100

So the sun has same probability of being even as it has of being odd

So probability of sum odd= 1/2

ANSWER 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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New post 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
Hence Answer is C.

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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New post 28 Dec 2017, 06:08
Bunuel wrote:
Mochad 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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New post 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

Final Answer:

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Re: A box contains 100 balls, numbered from 1 to 100. If three b   [#permalink] 11 Jan 2018, 16:47

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