4
$\begingroup$

A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd?

The answer to this question is 1/2. This is a a high level GMAT sample question and the working out in the answer shown is pretty long. When I picked 1/2 so quickly, the system assumed I guessed. My question is was it a lucky guess?

Or can we say

  • if we were to pick one ball instead of 3 the answer would have been 1/2?

  • same answer if we were to pick 2 balls?

If so, can we apply the same principle to 3 balls? Or is the rationale incorrect?

What about if we were to pick 4 or 5 balls?

Thank you.

$\endgroup$
  • $\begingroup$ Here is a restatement of your question that may help you think about it: A box contains 50 black balls and 50 white balls. You draw three of them. What's the probability that you get 1 or 3 white balls? The answer of 1/2 follows immediately from symmetry. Use this same restatement to address you other questions. $\endgroup$ – cardinal Apr 25 '13 at 1:57
  • $\begingroup$ The previous comment was meant to address the version where we sample without replacement. $\endgroup$ – cardinal Apr 25 '13 at 2:10
  • $\begingroup$ @cardinal The application of symmetry needs to be a little subtler when an even number of balls is drawn. BTW, a panel at the right of the review should show some statistics, beginning with a line of the form "Answers 1". $\endgroup$ – whuber Apr 25 '13 at 2:17
  • $\begingroup$ @whuber: Yes, I agree that the argument is subtler for even numbers. Maybe the wording of my comment didn't allude to that strongly enough; I was trying to leave something for the OP to think about. (Thanks for the tip, as well, which I was obviously unaware of!) $\endgroup$ – cardinal Apr 25 '13 at 2:24
6
$\begingroup$

This problem requires only one simple calculation; namely, that half the balls in the box are odd and half even.

One principle of random selection from a box is that only the proportion matters: the behavior (of sampling with replacement) of a box with 100 balls is the same as that with just two balls, one odd and one even, because in both cases exactly half the balls are odd and half even.

This problem enjoys a symmetry: if we were to relabel all the even balls as odd and all the odd balls as even, the proportions of each in the box would not change but the parity of the sum would change. Consequently, the chance of an odd sum must equal the chance of an even sum. Since that's all that can occur, each chance must be $1/2$.

Different reasoning is needed when drawing an even number of balls, for then the parity of the sum does not change after the relabeling. For example, if originally three odd balls and one even ball had been drawn (for an odd sum), then after relabeling, three even balls and one odd ball are drawn--again for an odd sum. One way to draw a conclusion in this case is to consider the situation just before the last ball is drawn. At this point an odd number of balls have come out, whence there is a $1/2$ chance that their sum is odd, $1/2$ that it is even. The final ball will either switch the parity (if it's odd) or not (if it's even), but whether it switches or not doesn't create any imbalance in the chances.

This is kind of interesting, because pursuing this argument (inductively) to its end leads to a curious situation. As before, suppose the box starts out with equal numbers of odd and even balls. However, after each step (when you draw one ball at random, observe its parity, and replace it), someone else adds to or removes from the box an arbitrary number of balls unknown to you. (All you know is that at least one ball remains in the box.) This changes the chances to completely unknown values, anywhere between $0$ and $1$. Let's further suppose that a third party, who has not observed these proceedings, gets to decide when you stop drawing, so you don't even know in advance how many balls ($n$) you will draw, except that it is nonzero. The parities of the final sum still have equal chances of $1/2$. The reason should now be obvious: the parity of the sum of the last $n-1$ balls could be anything; adding in the value of the first ball either changes the parity or not, with equal chances. As a result, there is a $1/2$ chance that the sum of all $n$ balls will be even and $1/2$ that it will be odd.

(This has practical applications in creating unbreakable ciphers, for instance.)

$\endgroup$
  • $\begingroup$ +1 Yes, using the symmetry is a much more elegant way of explaining this than the step by step approach in my answer. And closer to the intuition that "of course it's 1/2" because as you say, it has to be the same for both odds and evens, and if all the balls had their parity reversed, etc $\endgroup$ – Peter Ellis Apr 25 '13 at 2:25
  • $\begingroup$ (+1) Especially for the extension. $\endgroup$ – cardinal Apr 25 '13 at 2:27
  • $\begingroup$ @Cardinal I'm sure you see the cute theorem here about sums of coefficients of even powers in the expansion of $(1+x) \prod_i (1-a_i+a_i x)$, as well as the immediate proof. More generally we obtain a theorem about alternating coefficient sums of any polynomial having $-1$ as a root; or, vice versa, this trivial algebraic theorem immediately proves the extension. $\endgroup$ – whuber Apr 25 '13 at 2:52
3
$\begingroup$

I think it is very reasonable to quickly come to 1/2 by a similar process to what you suggest. Obviously if you pick one ball the answer is 1/2 as there are 50 odd numbers and 50 even to choose from. Now, assume you have picked one ball and hence have a equally chanced random odd or even number. After replacement you pick a second ball, also with an equally chanced random or or even number. Obviously the sum of those numbers has an equal chance of odd or even (because there are four equally likely combinations - odd+odd=even, odd+even=odd, even+odd=odd, and even+even=even). Now, you are back in the position you were after one ball - you have a number that if 50/50 even or odd. Regardless of how many balls you pick the process above works.

$\endgroup$
  • $\begingroup$ (+1) Using parity is the way to think about this. It also helps if, as you've done, you read the question carefully enough to notice it said with replacement. :-) I wish the review tab would have shown there was an answer because otherwise I wouldn't have written my (hasty) comment to the OP. Cheers. $\endgroup$ – cardinal Apr 25 '13 at 2:09
0
$\begingroup$

Actually, at the risk of restating what the other folks have said, I would just approach this problem by listing the eight outcomes when one picks three balls.

Definition: E,Even; O,Odd

Explanation:

The eight possibilities are as follows: OOO=O,OEO=E,OEE=O,EOO=E,OOE=E,EOE=O,EEO=O,EEE=E.

Therefore, from the eight possible outcomes, four possibilities produce an odd result. In other words, 4/8 = 1/2 resulting in the desired answer.

$\endgroup$
  • $\begingroup$ You are correct, but might be unaware that your answer could easily be misread as invoking an incorrect principle: namely, that the chances of individual outcomes must be equal. This "principle of indifference" is the source of many mistakes over a very long period of time. What is needed to shore up your argument, even though it's a trivial observation, is that because there are equal numbers of even and odd balls in the box, therefore the eight outcomes each have equal probabilities. (The axiomatic justification for this implication follows from the independence of the draws.) $\endgroup$ – whuber Apr 25 '13 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.