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

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

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).

Bunuel - can you please do this one
If it was without replacement?

so ill be sure i understood it the right way?

thanks.

Without replacement; the condition for getting odd doesn't change; only the probability of picking up the ball does;

OEE
EOE
EEO
OOO

50/100*50/99*49/98+50/100*50/99*49/98+50/100*49/99*50/98+50/100*49/99*48/98
=1/2*50/99*1/2+1/2*50/99*1/2+1/2*49/99*25/49+1/2*49/99*24/49

rest can be simplified.

Correct me if I am wrong, Bunuel.

Bunuel - u r amazing.great explanation from all aspects. thanks. +1
+1 fluke

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.

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.

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

Posted from my mobile device

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:

GMAT assassins aren't born, they're made,
Rich

Bunuel,
Need some help here. The question clearly states
what is the probability that the sum of the three numbers on the balls selected from the box will be odd ?

Now Either I choose EEO or OEE or EOE, in each case sum of three numbers is ODD.

Then Why should I consider the order in which ball is taken out ?
2even+1odd=odd
3odd=odd

These are the only 2 cases and there is only single box of balls.

Hi Harsh2111s,

These types of questions can often be approached in a couple of different ways, depending on whether you want to use "Permutations" or "Combinations" to get your answer. This approach uses Permutations.

You have correctly determined that for the SUM of the three numbers to be ODD, we either need to have 1 odd (among the 3 numbers) or 3 odds (among the 3 numbers) - and there are 4 possible ways for that to happen:

OEE
EOE
EEO
OOO

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 is an equal probability of any of those 4 results happening. For example:

(Odd)(Even)(Even) = (1/2)(1/2)(1/2) = 1/8

There are 4 equally-likely options, and we have to account for all of them: 4(1/8) = 4/8 = 1/2 probability of getting a sum that's odd.

GMAT assassins aren't born, they're made,
Rich

Say, the balls you took out are: 1st ball is 2, 2nd ball is 4 and 3rd ball is 5
This case is clearly different from: 1st ball is 5, 2nd ball is 4 and 3rd ball is 2

Let us understand why:
Take a different question: The balls you have are numbered 1 to 4. You need to pick 2 balls with replacement and get an odd sum. What is the probability?
Let us determine the total cases: We often use the logic that the number of ways of picking the first ball is 4 and that for the 2nd ball is 4
=> So total = 4 x 4 = 16 cases

Observe that as soon as we are doing 4 x 4, we are considering the order. Let's count:
11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44 => 16 cases

Thus, its natural, when we count the favorable cases, we must consider the order too:

Cases for odd: 12, 14, 21, 23, 32, 34, 41, 43 => 8 cases => p(odd) = 8/16 = 1/2
Cases for even: 11, 13, 22, 24, 31, 33, 42, 44 => 8 cases => p(even) = 8/16 = 1/2
When you pick 2 balls, the sum can either be even or odd. Thus, p(even sum) + p(odd sum) = 1 (satisfied)


If you wish to not consider the order, then the denominator will not be 16 either - finding the total cases as a selection is trickier than finding the cases as an ordered selection (arrangement)


Thus, coming back to our question too, we should consider the order.

You have 50 odd and 50 even balls.

Favorable cases:

Case 1: All 3 numbers are odd: 50 * 50 * 50 = 125 * 1000
Or
Case 2: 2 are even and 1 odd (EEO or EOE or OEE => 3 cases): favorable cases = 3 * (50 * 50 * 50) = 3 * 125 * 1000 = 375 * 1000

=> Favorable cases = 125 * 1000 + 375 * 1000 = 1000 * 500

Total cases = 100 * 100 * 100 = 1000 * 1000

=> Probability = (1000*500) / (1000*1000) = 1/2


Alternative approach:

Case 1: All 3 numbers are odd: probability = 50/100 * 50/100 * 50/100 = 1/8
Or
Case 2: 2 are even and 1 odd (EEO or EOE or OEE => 3 cases): probability = 3 * (50/100 * 50/00 * 50/100) = 3/8

=> Probability = 1/8 + 3/8 = 4/8 = 1/2

Option C

Dear IanStewart GMATGuruNY MathRevolution VeritasKarishma Bunuel GMATinsight,

MODIFIED QUESTION: A box contains 3,000 balls, numbered from 1 to 3,000. If NINETY NINE balls are selected at random and WITH replacement from the box, what is the probability that the sum of the NINETY NINE numbers on the balls selected from the box will be odd?

Q1. Will the answer to the above problem still be 1/2 because of symmetry in the number of odds and evens in the pool we pick?

Q2. If the question were WITHOUT replacement, is there a logical approach to solve to above modified question within time limit?

Yes. If your sum is even after 98 selections, you need to pick an odd number, and if your sum is odd after 98 selections, you need to pick an even number. Either way, you always have a 1/2 probability of picking the "right" type of number with your 99th pick, so 1/2 is the answer.





Yes, but I think that situation is probably too complicated for the GMAT. The answer is again 1/2;intuitively there should be no reason an odd or even sum should be more likely than the other if the set is half even, half odd. There are a few ways to prove that's the answer, though some rely on math beyond the scope of the test. Using only GMAT-level math, you could notice that these sequences have an equal probability of occurring:

EEE...EEE (all 99 selections are even) -- even sum
OOO...OOO (all 99 selections are odd) -- odd sum

and so do these sequences:

EEE...O... EEE (98 even selections, one odd) -- odd sum
OOO...E...OOO (98 odd selections, one even) -- even sum

and so on. So the sequences producing odd sums are balanced out by equally likely sequences producing even sums, and the answer is 1/2.

More abstractly, if you take any sequence of 99 Evens and Odds, and then just flip all the Even things to Odds, and all the Odds to Even, you'll get an equally likely sequence with the opposite type of sum (for example if you have a sequence with 46 evens and 53 odds, that produces an odd sum, but change all the even numbers to odd ones and vice versa, and now you get an even sum). So an even sum and odd sum must be equally likely.

Still, that question seems beyond GMAT scope, in part just because the numbers are so big that direct computational approaches become impossible. Usually in GMAT questions like this, there will be a 'fast' conceptual approach (like the ones I explained above) available, along with a slower-but-practical calculational approach (using ordinary probability rules). If you had a set of 10 numbers and 3 selections without replacement, that would be a more GMAT-like scenario.