Why is Binomial Probability used here?

A test consists of 10 multiple choice questions with five choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions.
What is the probability of getting exactly 6 questions correct on this test?

The answer says:

$$P = \binom{10}{6} (0.2)^6 (0.8)^4$$

I was thinking though:

$$P(A \text{and} B) = P(A) \cdot P(B)$$

Probability of getting $6$ right and $4$ wrong would be just:

$$(0.2)^6 (0.8)^4$$

Why is the binomial coefficient used there?

Solutions Collecting From Web of "Why is Binomial Probability used here?"

heropup gave a really good answer, but let me expand on this a bit for you to give you a bit of a deeper understanding from a different perspective.

We know that the total probability for any event happening for all cases must be $1$. If the problem was solved the way you presented, would the total probabilities add up to $1$? Clearly not.

Let’s look at each individual question, which by itself has probabilities that add up to one. For a particular question, getting it right is a $0.2$ chance and getting it wrong is a $0.8$ chance, a total of $1$. If we go question by question, we take the probability to a power, so let’s do that with the total probability:

(0.8 + 0.2) &= 0.8 + 0.2 \\
(0.8 + 0.2)^2 &= 0.8^2 + 2 (0.8)(0.2) + 0.2^2 \\
(0.8 + 0.2)^3 &= 0.8^3 + 3 (0.8)^2(0.2) + 3 (0.8)(0.2)^2 + 0.2^3 \\
(0.8 + 0.2)^4 &= 0.8^4 + 4 (0.8)^3(0.2) + 6 (0.8)^2(0.2)^2 + 4 (0.8)(0.2)^3 + 0.2^4

And so on and so forth. This is a direct application of the binomial theorem:
(x+y)^n = \sum_{k=0}^n {\binom{n}{k}x^{n-k}y^k}
for $x = 0.8$ and $y = 0.2$.

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 $$\begin{align*} (R, R, R) \\ (R, R, W) \\ (R, W, R) \\ (R, W, W) \\ (W, R, R) \\ (W, R, W) \\ (W, W, R) \\ (W, W, W) \end{align*}$$ 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, $\binom{n}{k}$ arrangements of $k$ correct answers out of $n$ questions, and each distinct ordering of correct/incorrect answers is a distinct elementary outcome.

You’ve now calculated the probability that you are first right 6 times and then wrong 4 times. But you could of course also have a different combination like right 5 times, then wrong 4 times and then right again. The binomial coefficient gives you the amount of combinations you could have.