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Probability GMAT Prep

by dhonu121 » Fri Jun 15, 2012 7:06 am
The following problem is from the MBA.com's preparation CD.

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

I have a problem with the solution of this problem.
Why is order considered here ? We just want to know the probability of getting the sum odd.
Thus we have to either get all three numbers odd or (one odd and two even.) Why is ordering put into picture in the second case is beyond my understanding.

Can somebody please help ?
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by voodoo_child » Fri Jun 15, 2012 7:16 am
dhonu121 wrote:The following problem is from the MBA.com's preparation CD.

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

I have a problem with the solution of this problem.
Why is order considered here ? We just want to know the probability of getting the sum odd.
Thus we have to either get all three numbers odd or (one odd and two even.) Why is ordering put into picture in the second case is beyond my understanding.

Can somebody please help ?
P(o) = P(e) = 1/2

There are four combinations out of 8 for which the probability of sum = odd.

HEnce, it should be C)

HTH
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by voodoo_child » Fri Jun 15, 2012 7:18 am
btw, there is a good treatment here :

https://www.manhattangmat.com/forums/a-b ... t7854.html


There are some other methods as well... HTH
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by dhonu121 » Fri Jun 15, 2012 7:39 am
voodoo_child:Neither your post here nor your link here was of any help.
I would again emphasize my problem: Why is ordering taken into account here when balls are drawn with replacement ?
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by voodoo_child » Fri Jun 15, 2012 7:55 am
dhonu121 wrote:voodoo_child:Neither your post here nor your link here was of any help.
I would again emphasize my problem: Why is ordering taken into account here when balls are drawn with replacement ?
Sure - I can help you. First I want to see your solution. Can you please post your solution so that I can see where you are going wrong.
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by GMATGuruNY » Fri Jun 15, 2012 7:56 am
dhonu121 wrote:The following problem is from the MBA.com's preparation CD.

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

I have a problem with the solution of this problem.
Why is order considered here ? We just want to know the probability of getting the sum odd.
Thus we have to either get all three numbers odd or (one odd and two even.) Why is ordering put into picture in the second case is beyond my understanding.

Can somebody please help ?
Whenever we multiply probabilities, the resulting product represents the probability that ONE PARTICULAR ordering of the events will happen.
P(RRG) = the probability that the first two outcomes are R and the third outcome is G.
P(AABBCC) = the probability that the first two outcomes are A, the next two outcomes are B, and the last two outcomes are C.
Whether the selections are made with replacement or without doesn't matter: the resulting product represents the probability that ONE PARTICULAR ordering of the events will occur.
When there is more than one possible ordering -- as in the problem at hand -- we must account for ALL of the different orderings that will yield a favorable outcome.

P(exactly one odd):
P(OEE) = 1/2 * 1/2 * 1/2 = 1/8.
Thus, the probability that the FIRST number is odd and the remaining numbers are even = 1/8.
But this result accounts for only ONE WAY to select exactly one odd.
There are TWO OTHER WAYS to select exactly one odd: EOE and EEO.
To account for the other two ways to select exactly one odd --for a total of THREE ways to select exactly one odd -- we multiply the result above by 3:
P(exactly one odd) = 3 * 1/8 = 3/8.

For more about this problem, check my posts here:

https://www.beatthegmat.com/probability-t94780.html

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by dhonu121 » Fri Jun 15, 2012 8:49 am
Thanks for the elaborate explanation Mitch.
However, we are asked the probability of getting the SUM odd.
Now, once we have taken out the balls, 1st then 2nd and then 3rd, the SUM should be odd.
What you are doing here is finding the probability of getting one odd in three drawings, but in effect, we are supposed to find the SUM odd, not number of ways of getting 1 odd.
In what order were the balls drawn is not at all coming into picture. Why is then we are considering order here.
Were we supposed to place the place in a horizontal or vertical line and then calculate the number of ways of getting odd, then counting various permutation would have made sense. But here,sorry , I am not able to get a hang of it.
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by GMATGuruNY » Fri Jun 15, 2012 9:24 am
dhonu121 wrote:Thanks for the elaborate explanation Mitch.
However, we are asked the probability of getting the SUM odd.
Now, once we have taken out the balls, 1st then 2nd and then 3rd, the SUM should be odd.
In what order were the balls drawn is not at all coming into picture. Why is then we are considering order here.
Were we supposed to place the place in a horizontal or vertical line and then calculate the number of ways of getting odd, then counting various permutation would have made sense. But here,sorry , I am not able to get a hang of it.
Let's say 3 people -- A, B and C -- each hand you a slip of paper on which is written one of the numbers between 1 and 100, inclusive.
You now hold a COMBINATION of 3 slips of paper in your hand.
You want the sum of the numbers to be odd.

Case 1: The sum will be odd if all 3 numbers are odd.
There is only ONE WAY for this outcome to occur:
A = odd, B = odd, C = odd.
Thus, P(OOO) = 1/2 * 1/2 * 1/2 = 1/8.

Case 2: The sum will be odd if exactly one of the slips of paper holds an odd number, while the other two each hold an even number.
Here, there are THREE ways to get a good outcome:
A = odd, B = even, C = even
A = even, B = odd, C = even
A = even, B = even, C = odd.
Even though you hold in your hand ONE COMBINATION of 3 numbers, there are still THREE WAYS for this combination to contain one odd number and 2 even numbers.
Thus, you must account for all 3 ways that this combination of numbers could yield one odd number and two even numbers:
P(A is odd, B is even, C is even) = 1/2 * 1/2 * 1/2 = 1/8.
PIA is even, B is odd, C is even) = 1/2 * 1/2 * 1/2 = 1/8.
P(A is even, B is even, C is odd) = 1/2 * 1/2 * 1/2 = 1/8.
Since either the first way OR the second way OR the third way will yield a good outcome, add the fractions:
P(one odd and two evens) = 1/8 + 1/8 + 1/8 = 3/8.

Since either Case 1 OR Case 2 will yield a favorable outcome, add the results above:
1/8 + 3/8 = 1/2.
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by dhonu121 » Fri Jun 15, 2012 11:53 am
got it !!
Thanks !
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by ziko » Mon Jun 18, 2012 12:43 am
Let me give my understanding as well, i would be happy if it helps.
So no matter how many balls we have there are equal number of even and odd numbers, and they are picked with replacement (which is important to know). We have three picks with 2 possibilities for each (even odd even odd even odd) 6C3 overal 20 possible pairs. Our favorable outcomes are even+even+odd (in any order) OR odd+odd+odd, so the combinations are 3C2*3C1=9 and 3C3=1 accordingly, overal 10. Therefore, sum of three numbers will be odd in 10 out of 20 cases, basically 1/2. (C)
Hope that helps!
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by Kaustubhk » Thu May 25, 2017 9:49 am
Hi Mitch,

The doubt which dhonu121 had asked of ordering of the numbers. Does the catch lie in the question itself?

The last part clearly states that the 'what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Since the probability of sum of three numbers on the ball been odd is asked we need to consider every probable event which is possible in making the sum odd.

Correct me if i'm wrong.
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by [email protected] » Thu May 25, 2017 10:07 am
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: C

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by Matt@VeritasPrep » Fri May 26, 2017 2:22 pm
Kaustubhk wrote:Hi Mitch,

The doubt which dhonu121 had asked of ordering of the numbers. Does the catch lie in the question itself?

The last part clearly states that the 'what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Since the probability of sum of three numbers on the ball been odd is asked we need to consider every probable event which is possible in making the sum odd.

Correct me if i'm wrong.
You're right, yup! We need to consider two scenarios (exactly one odd and exactly three odds), and one of those scenarios is order-dependent: exactly one odd has to be arranged OEE, EOE, and EEO.
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