Monday, November 7, 2011

My Take on the Best Statistics Question Ever

A difficulty that students often face is that exam questions like this may be interpreted a number of ways, with each interpretation leading to a distinctly different answer. In this post, I offer two such interpretations.

The first leads to the same answer as one of our readers: "0% the correct answer is not among the choices." Here we assume that the question that needs a correct answer is "what is the chance you will be correct," and that you would only be "correct" if the probability of picking your answer is the same as the value of your chosen answer.

Say, you randomly pick "25%". Since two out of the four choices represent "25%", then the probability of randomly picking it is 50% (assuming equal likelihood). And since your answer--25%--is not the same as the probability of picking your answer--50%--then you will have been incorrect.

Using the same logic, it's easy to see how picking the other two available choices, 50% and 60%, would also be incorrect. This means that the chance that you will be correct is zero--there is no chance that you will be right!

Of course, if you read the problem differently, you should arrive at a different answer. For example, if we assume that "being correct" refers to some other arbitrary question, that the choices pertain to that question and not to the chance that your answer will be correct, and that one of the given choices is the correct answer, then we'll arrive at an altogether different solution.

We are still interested in the probability of randomly picking a correct answer, but this time we do not limit this probability to the given choices. We could go through equations and basic laws of probability to solve the problem, and in the end we'll most probably arrive at the correct solution, but there is a simple way of reasoning out the correct solution instead. 

First we recognize that there are only three possible answers to the problem: 25%, 50%, and 60%. If we assume that the three are equally likely to be the correct answer, then the probability that each will be correct is 1/3. Or,
  • If C = the correct answer, where C = 25%, 50%, or 60%. Assuming these choices are equally likely to be correct, then the probability of C, P(C = 25%) = P(C = 50%) = P(C = 60%) = 1/3.

Since the probabilities are all the same, then the actual choice we pick does not matter: whatever we choose, the probability that it will be the correct answer is 1/3.

Can you think of another approach to solving the problem? Feel free to share your thoughts in the comments section below.

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