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705-805 Level|   Probability|            
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A box contains 100 balls, numbered from 1 to 100. If three b 18 Jun 2020, 05:06
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varotkorn

However, if we chose ONE HUNDRED (instead of 99), then the odd sums would NOT balance equally with even sums, right?

Let's say I have:
EEE...EEE (all 100 selections are even) -- even sum
OOO...OOO (all 100 selections are even) -- still even sum

These 2 scenarios, however, have the same probability of occurring as you mentioned above.
Show more

Changing it to an even number of selections, without replacement, makes an out-of-scope problem even harder. Then you can't take advantage of symmetry any more, and the possibilities don't balance out perfectly. The answer will be very close to 1/2, but I don't immediately see a way to find the precise answer quickly using only GMAT-level math. You could certainly see a similar problem with, say, 10 or 20 balls in total, and 2 or 4 selections in total, since those problems are tractable using standard probability rules, but I wouldn't worry about a situation like the one you're asking about.
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A box contains 100 balls, numbered from 1 to 100. If three b 26 May 2021, 23:49
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Solution:

If a+ b+ c=odd

=>Case 1- Either all are odd OR

Case 2-a, b are even and c is odd in the order of a,b,c being picked

(a)Even +(b)Even + (c)Odd=Odd)

P(Case 1) = (1/2)^3

P(Case 2) = 1/2 * 1/2*1/2 * 3 (To count the cases of abc, acb, or cab)

=>Probability = 1/8 +3/8 =4/8=1/2 (option c)

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A box contains 100 balls, numbered from 1 to 100. If three b 26 Jun 2021, 07:35
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Bunuel
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.

Answer: C.

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.
Show more


Why does order matter here for EEO

We are anyway replacing the ball taken out.

Why does it matter if i were to replace anyway.

443 344 and 434 is the same ? What am i missing in this logic ?

Please help

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hD13

Why does order matter here for EEO

We are anyway replacing the ball taken out.

Why does it matter if i were to replace anyway.

443 344 and 434 is the same ? What am i missing in this logic ?
Show more

In this specific question, you'll often end up with the correct answer, 1/2, even if you don't think order matters, or if you don't notice we're doing things with replacement. But in other similar questions, you'd get the wrong answer if you did that, which I'll illustrate with a couple of simple examples:

If two numbers will be chosen from the set {1, 2} without replacement, what is the probability their sum is odd?

Here, if we choose without replacement, we automatically pick 1 and 2, and the sum is always 3, so the probability the sum is odd is 100%. Contrast that with the question

If two numbers will be chosen from the set {1, 2} with replacement, what is the probability their sum is odd?

There are simpler ways to do the question, but one way to do this is to list all the things that can happen. If we pick '1' first, there will be two numbers we can pick next, and same if we pick '2' first, so there will be four sequences we can pick in total:

1, 1
1, 2
2, 1
2, 2

and in two out of four cases, the sum is odd, so the answer is 2/4 = 1/2. This illustrates why order matters: here, there are two different ways to get a 1 and a 2, in some order, but only one way to get a 1 and a 1. It's twice as likely that we pick 1 and 2 (in some order) than that we pick 1 and 1, and we need to be sure to account for that. If instead we thought "1+2 and 2+1 are the same, so that's only one event", we'd think there are only three events in total (1,1 and 1,2 and 2,2), one of which gives an odd sum, and we'd get the incorrect answer 1/3.

In the particular problem in this thread, the other reason we care that selections are done with replacement is because that's what the question tells us is happening here. There's a reason for that - it's a much simpler problem this way, and the GMAT likes simple problems. If you instead solve the problem without replacement, no matter how you solve, the math turns out to be messier, though it turns out you'll still get the same answer in the specific situation the question outlines.
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Why we can not solve in reverse way. I.E. selecting probability of picking an even no and substracting from 1.
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A box contains 100 balls, numbered from 1 to 100. If three b 13 Aug 2021, 19:39
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Hi sanjeevsinha082,

You CAN approach this question in that way - but doing so would probably take you the same amount of time.

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 Odd numbers and Even numbers, each of these 8 options has the same 1/8 probability of happening. If you want to put together the list of the options that total an EVEN result, then they would be....

EEE
OOE
OEO
EOO

Four of the eight options. 4/8 = 1/2. We can then subtract that from 1...

1 - 1/2 = 1/2

...and that is the probability of having a sum that is ODD.

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A box contains 100 balls, numbered from 1 to 100. If three b 07 Jun 2023, 10:35
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Bunuel
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.

Answer: C.

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.
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Shouldn't we also multiply 3 with 2*1/2 as as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways??
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Ahanasarkar99
Bunuel
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.

Answer: C.

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.


Shouldn't we also multiply 3 with 2*1/2 as as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways??
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Not sure I understand what you mean. EEO can occur in three ways: EEO, EOE, OEE, hence we multiply (1/2)^2*1/2 by 3 as shown in the solution.
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Bunuel
mbafall2011
The order will matter if there is no replacement:

If the first pick is even, the probability of a second even will be 49/99 and odd will be 50/99.

Also, im looking at these as mutually independent events rather than Probability of EEO +EOE etc.

But if i write out all the possibilities

ooo
ooe
oeo
oee
eoo
eoe
eeo
eee

then i can see that 4 out of 8 picks are favorable.

This one is tricky!

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.

Next, the way you are doing (the red part) is correct only for the cases in which there are equal # of even and odd numbers (for example if there were balls numbered from 1 to 99 this approach wouldn't be corrorect, so after all the probability approach is better).
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"if there were balls numbered from 1 to 99 this approach wouldn't be corrorect" - why is the solution given only eligible in which the 2 types of balls even/odd are same in number. Even if they are different in number, the number of possibilities of whether a ball is even or odd stays the same i.e. 1/2. (Also, I am assuming this is a binomial probability question. Is my thinking correct ?)

Posted from my mobile device
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A box contains 100 balls, numbered from 1 to 100. If three b 12 May 2024, 05:09
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­Classic probability we all love so much:

­
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odd+ odd+odd=odd : (p)=(1/2)^3 = 1/8
Even+even+odd=odd : (p)= (1/2)^2 * (1/2) * (3!/2!)= 3/8
(above arrangment of even even odd situation 3!/2!)
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A box contains 100 balls, numbered from 1 to 100. If three b 04 Apr 2025, 02:35
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SmitKhurana
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
Show more
The sum of 3 will be odd when either all are odd or two are even and one is odd

Probability (all odd) = (1/2)*(1/2)*(1/2) = 1/8

Probability (One odd and two even) = (1/2)*(1/2)*(1/2)*3C2 = 3/8

Required probability = (1/8) + (3/8) = 1/2

Answer: Option C
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A box contains 100 balls, numbered from 1 to 100. If three b 19 Aug 2025, 20:42
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Why are we not accounting for the multiple ways of getting EEO or OOO?

For instance 2,4,5 is different from 4,2,5

3,5,7 is different from 5,7,3
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dc48048
Why are we not accounting for the multiple ways of getting EEO or OOO?

For instance 2,4,5 is different from 4,2,5

3,5,7 is different from 5,7,3
Show more

Please review the discussion. The case with two evens and one odd is multiplied by 3 exactly because it can occur in three different ways: EEO, EOE, or OEE. The OOO case isn’t multiplied by anything because there’s only one way for all three picks to be odd.
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A box contains 100 balls, numbered from 1 to 100. If three b 15 Sep 2025, 19:41
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mbafall2011
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.
Show more

22% people mark option A for this one here. That’s a huge number. Many people have explained as well as or ever better than I could explain why EEO, EOE and OEE need to be considered different.

I’ll share an additional thought.

The sum of the three balls can only be odd or even.

If we think that the probability of the sum being odd is 1/4, the probability of the sum being even would have to be 3/4.

i. At this point, you could check mathematically what you get the probability for the sum to be even as. To get an even sum, we’d need to draw three even numbered balls or one even and two odd numbered balls. If you go about calculating the probability for an even sum the same way as you did for an odd sum, then you’d get 1/4 again. 1/4 probability that the sum is odd, 1/4 that the sum is even. Then what’s happening in the remaining 1 - 1/4 - 1/4 = 1/2 probability? The sum can’t be anything other than odd or even. That means there’s a mistake.

ii. We could also think about it logically. 1 - 100: there are equal number of odd and even numbered balls. The probability of the sum being odd or even should be equal. If I’m getting my answer as 1/4, I can ask: why is the probability so skewed toward the sum being even (3/4)? That could indicate to me that I could have made a mistake somewhere.

- -

This logic be a way to solve the question too.

i. There’s an equal number of odd and even numbered balls.
ii. The sum can ONLY be odd or even. No third thing.
iii. The sums of odd and even are equally likely. (I could enumerate this too: For an odd sum: either all three are odd, or two even and one odd. For an even sum: either all three are even, or two odd one even.)
iv. Two possible outcomes. Both equally likely. So, the probability of each would be 1/2.
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Let me help you crack this probability question efficiently. The trick here is recognizing this is fundamentally about odd/even patterns, not specific numbers.

Core Addition Rules to Remember:

Odd + Odd = Even
Even + Even = Even
Odd + Even = Odd

Step 1: When is a sum of three numbers odd?
Let's work through the possibilities systematically:

Odd + Odd + Odd = (Odd + Odd) + Odd = Even + Odd = Odd ✓
Odd + Odd + Even = (Odd + Odd) + Even = Even + Even = Even ✗
Odd + Even + Even = Odd + (Even + Even) = Odd + Even = Odd ✓
Even + Even + Even = Even + Even = Even ✗

Key finding: The sum is odd when we have either all three odd OR exactly one odd.

Step 2: Calculate individual probabilities
From 1 to 100:

Odd numbers: 1, 3, 5, ..., 99 → Total = 50
Even numbers: 2, 4, 6, ..., 100 → Total = 50

Therefore:
\(P(\text{Odd}) = \frac{50}{100} = \frac{1}{2}\)
\(P(\text{Even}) = \frac{50}{100} = \frac{1}{2}\)

Step 3: Calculate probability for each favorable scenario

Case 1: All three odd (OOO)
\(P = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

Case 2: Exactly one odd
This can happen in 3 ways:

OEE: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)
EOE: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)
EEO: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

Total for Case 2: \(\frac{3}{8}\)

Step 4: Final Answer
Total probability = \(\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}\)
Answer: C

You can check out the complete solution on Neuron by e-GMAT, which includes advanced pattern recognition techniques and time-saving strategies that work across similar problems. You'll also discover why many test-takers incorrectly choose 3/8 and how to avoid this trap consistently.

Feel free to access detailed solutions for many other similar official questions with practice quizzes tailored to your weaknesses.
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Hi Sir,

Here they are asked the sum of 3 random numbers, then why we have to multiply the probability of even or odd instead of adding

Can you please explain


Bunuel
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.

Answer: C.

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.
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adipisciet
Hi Sir,

Here they are asked the sum of 3 random numbers, then why we have to multiply the probability of even or odd instead of adding

Can you please explain



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The reason we multiply is because we want the probability of several events happening together in a row (for example, first Even, then Even, then Odd). To get the probability of all three happening together, we multiply their chances.

The reason we add is because there are different possible sequences (EEO, EOE, OEE). Each sequence is a separate way to reach the same outcome, so their probabilities are added.
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