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

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

VeritasKarishma chetan2u MathRevolution IanStewart

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.

Why we can not solve in reverse way. I.E. selecting probability of picking an even no and substracting from 1.

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.

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

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??

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.

"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 ?)

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