# Probability, does the order matter?

According to Webster’s New Collegiate Dictionary, a divining rod is “a forked rod believed to indicate [divine] the presence of water or minerals by dipping downward when held over a vein.” To test the claims of a divining rod expert, skeptics bury four cans in the ground, two empty and two filled with water. The expert is led to the four cans and told that two contain water. He uses the divining rod to test each of the four cans and decide which two contain water.

a) List the sample space

b) If the divining rod is completely useless for locating water, what is the probability that the expert will correctly identify (by guessing) both of the cans containing water?

W = water, E = empty

For a) I have, (W1, E1), (W1,E2), (W1,W2), (W2, E1), (W2, E2), (E1, E2), but I'm just wondering if order matters. Is (E1, W1) different from (W1, E1)?

b) is it just 1/6?

• No, order does not matter. The person is selecting two cans, not a distinct "Can 1" and "Can 2". Your answer for $b)$ is also correct, because each element in the sample space is equally likely to occur. – Eric Tressler Sep 17 '15 at 21:16
• I don't like Question a). But some textbook writers seem to like this sort of thing, since it occurs distressingly often. The sample space consisting of the $\binom{4}{2}$ choices of two cans is indeed an appropriate sample space, arguably the best one for solving Question b). But one could also use the sample space of all sequences of two choices. This sample space has $12$ elements, of which $2$ are "favourable." – André Nicolas Sep 17 '15 at 21:25
• I disagree slightly with @AndréNicolas in that, while a) is not a very useful question in general, it might be a good notion to iron out in an introductory probability course (it's better not to leave it as a fuzzy concept). – Eric Tressler Sep 17 '15 at 21:32
• @EricTressler: A question like "List the sample space you are using to describe the experimental outcomes" is fine. To my mind, list the sample space is not. – André Nicolas Sep 17 '15 at 21:35
• @AndréNicolas Okay. I can't argue with that. – Eric Tressler Sep 17 '15 at 21:37

Everything you have said is correct, but I want to elaborate a little bit on $b)$. There are four cans, two of which are being chosen essentially at random. There are $$\binom{4}{2} = 6$$ possible choices of two cans, and only one of those choices results in two cans of water. Since each pair of cans is equally likely to be chosen, there is a $1/6$ probability that the two cans containing water are chosen.