Step 1

The short answer is that order matters.

The long answer begins with an example. Suppose you had only 3 questions, and each question was True/False, so you have equal probabilities of answering any given question correctly by random guessing. Then you can easily enumerate the possible outcomes of whether or not you get Questions 1, 2, or 3 correct. Let the ordered triplet $({q}_{1},{q}_{2},{q}_{3})$ represent the outcomes of questions 1,2,3 in that order, where ${q}_{i}\in \{R,W\}$ where R indicates a right answer, and W indicates a wrong answer. Then your outcomes are

$$\begin{array}{r}(R,R,R)\\ (R,R,W)\\ (R,W,R)\\ (R,W,W)\\ (W,R,R)\\ (W,R,W)\\ (W,W,R)\\ (W,W,W)\end{array}$$

and now it is obvious that there are 8 elementary outcomes. If I asked you what the probability is of getting exactly 2 correct answers out of three, you would say 3/8. But your method of reasoning for the original question would have given $(1/2{)}^{3}=1/8$, which is clearly not the case.

So why is your method not correct? Because there are, in general, $(}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )$ arrangements of k correct answers out of n questions, and each distinct ordering of correct/incorrect answers is a distinct elementary outcome.

###### Did you like this example?

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