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Probability

by anks17 » Thu Nov 03, 2011 6:27 pm
Please answer the following question from GMAT Prep with explanation:-

A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?

a)3/8
b)1/2
c)5/8
d)1/4
e)3/4
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by fcabanski » Thu Nov 03, 2011 8:16 pm
This is with replacement, and there are an equal number (50) of even and odd numbers. That means any of these combinations is equally likely.

If the first ball selected is even, the possibilities are as follows. An X indicates that the resulting sum would be odd:

EEE
EEO x
EOE x
EOO

If the first ball selected is odd, the possibilities are as follows. An X indicates the resulting sum would be odd:

OOO x
OOE
OEO
OEE x

There are 8 possible and equally likely outcomes, 4 of them result in an odd sum. The probability is 4/8 = 1/2.


Also this can be done via formula, again recognizing that evens and odds are equally likely.

The probability of the first ball being even is 1/2 times the probability of the second ball being even 1/2 times the probability of the third ball then being odd 1/2 = 1/2 * 1/2 * 1/2 = 1/8

OR

The probability of the first ball being even 1/2 times the probability of the second ball being odd 1/2 times the probability of the third ball being even 1/2 = 1/2 * 1/2 * 1/2 = 1/8

If all three balls are even, or if two are odd, the sum will be even. So that's it when the first ball is even.

If the first ball is odd it's:

OR

1/2 times second ball odd 1/2 times third ball odd 1/2 = 1/2 * 1/2 * 1/2 = 1/8

OR

1/2 times second ball even 1/2 times third ball even 1/2 = 1/2 * 1/2 * 1/2 = 1/8

That's it when the first ball is odd.

P (A or B or C or D) - PA + PB + PC + PD = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2
Last edited by fcabanski on Thu Nov 03, 2011 8:25 pm, edited 1 time in total.
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by saketk » Thu Nov 03, 2011 8:19 pm
anks17 wrote:Please answer the following question from GMAT Prep with explanation:-

A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?

a)3/8
b)1/2
c)5/8
d)1/4
e)3/4
100 balls with replacement -- means every time you can choose a ball in 100 ways

So total number of ways = 100*100*100

There are equal number of Odd and even numbered balls in the bag.

Every time you have 50 ways to choose an even or odd ball = 50*50*50

But the constraint is -- sum should be ODD at the end.

Either you can have a even numbered ball or odd numbered ball in every selection.

i.e. 2 possibilities per selection -

Total ways = 2*2*2 = 8 .. since there are equal number of odd and even balls. You will have exactly 4 possibilites when the sum is 0dd.

So the final answer is

[50*50*50]*4 / [100*100*100] = 1/2
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by pemdas » Thu Nov 03, 2011 8:42 pm
the probability the ball is selected at any time is 1/100, the probability that the ball selected is odd is 1/2 and even is 1/2. Since the effective probability for our answer choice (odd or even) should not account for the amounts but rather for the odd/even we ignore 1/100 ;) For the sequence of three selections in one set the probability will be (1/2)^3 or 1/8

the sum of three numbers can be odd, if [two numbers are even and one is odd, (1/2)^3] Plus [three numbers are odd, (1/2)^3]. No need for the selections of three out of one hundred.
1/8 + 1/8
the above makes 1/4

now that we have only two sets resulting in odd out of [one even and two odd, C(3,2) ways] & [all three even, 1 way]& [all three odd, 1 way] & [two even, one odd, C(3,2) ways]. Total makes 3+1+1+3=8 in the denominator and (3+1=4) in the numerator hence 4/8 makes 1/2

I have come with two answers based on two alternative solutions.
anks17 wrote:Please answer the following question from GMAT Prep with explanation:-

A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?

a)3/8
b)1/2
c)5/8
d)1/4
e)3/4
Last edited by pemdas on Thu Nov 03, 2011 8:55 pm, edited 1 time in total.
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by satishchandra » Thu Nov 03, 2011 8:48 pm
We have 50 even and 50 odd integers in between 1 to 100.

In order to get sum an odd number, we can have 3 ODD numbers or 2 EVEN+1 ODD number.

Probability with replacement = 50*50*50/(100*100*100)+50*50*50/(100*100*100) = 1/4

Answer : D

Whats the OA?
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by pemdas » Thu Nov 03, 2011 9:10 pm
it's actually interesting how GMAT Prep question tests our knowledge of independent probabilities by allowing first sampling with replacement out of 100 and then suggesting that the balls are numbered from 1 to 100, finally asking odd/even which puts us right in the vision of 1/2 "fair coin" case which is false. Because we may not leave some numbers in the pool and our probability condition will change from 1/2 to something else, we must account somehow for 1/100 selections too, but we are suggested that there's replacement rule so we leave it out here.

next, we have (1/2)^3 possibility for each of the three balls to be equally odd or even; it's just 1/2 and 1/2 and 1/2 all the time. We need to operate with the selections resulting in odd out of the total possible selections. My final answer is 1/2.
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by shankar.ashwin » Thu Nov 03, 2011 9:16 pm
I prefer doing this intuitively, not sure if the answer is correct though.

List the possibilities of getting an even sum or an odd sum.

For Even = (Even) + (Even) + (Even)
(or) Even = (Odd) + (Odd) + (Even)

The arrangement of getting odd and even does not matter here. So you got only 2 possibilities.

Same way for Odd;

Odd = (Odd) + (Odd) + (Odd)
(or) Odd = (Odd) + (Even) + (Even).

Again you have only 2 possibilities. (Order of getting them does not matter here)

So IMO you got equal chances of getting an odd or an even sum. So probability should be [spoiler]1/2[/spoiler]
Could you confirm the OA?
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by saketk » Thu Nov 03, 2011 9:17 pm
satishchandra wrote:We have 50 even and 50 odd integers in between 1 to 100.

In order to get sum an odd number, we can have 3 ODD numbers or 2 EVEN+1 ODD number.

Probability with replacement = 50*50*50/(100*100*100)+50*50*50/(100*100*100) = 1/4

Answer : D

Whats the OA?
There is a flaw in your solution -- you forgot to consider one more case..

ODD+ODD+ODD.

Answer should be 1/2. Scroll up 2 posts..
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by satishchandra » Thu Nov 03, 2011 9:24 pm
There is a flaw in your solution
Absolutely right. There was flaw in my approach.

I corrected it. Possibly the simplest approach is as under.
The all possible cases can be:
{EEO,EOE,OEE,OOO}
E-Even
O-odd
= (1/2*1/2*1/2)+(1/2*1/2*1/2)+(1/2*1/2*1/2)+(1/2*1/2*1/2) = 1/2
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by GMATGuruNY » Fri Nov 04, 2011 3:48 am
anks17 wrote:Please answer the following question from GMAT Prep with explanation:-

A bag contains 100 balls numbered from 1 to 100. If 3 balls are selected randomly from the bag with replacement, what is the probability that sum of the numbers on 3 balls is odd ?

a)3/8
b)1/2
c)5/8
d)1/4
e)3/4
Since there are 50 odd numbers and 50 even numbers:
P(odd number) = P(even number) = 1/2.

Ways to get an ODD sum:
3 odd numbers
1 odd number, 2 even numbers

Ways to get an EVEN sum:
3 even numbers
1 even number, 2 odd numbers

Since P(odd) = P(even) = 1/2:
P(3 odds) = P(3 evens).
P(1 odd, 2 evens) = P(1 even, 2 odds).

Thus:
P(odd sum) = P(even sum) = 1/2.

The correct answer is B.
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