understanding the binomial distribution

Question

I'm reading a chapter on bernoulli and binomial distribution and there is an exercise which reads:

Assume you are taking a multiple choice quiz (10 questions) and each question has 4 choices. Since you are not prepared for the quiz, you randomly choose an answer for each question independent of other questions...

Then it asks for the probability that you answer the first one right and the second one wrong.

In the solution it starts with the probability we answer every question wrong and the answer is given as:

$$({3\over4})^{10}$$

Then it says the probability that we answer one question correctly is:

$${1\over4}\times({3\over4})^{9}\times10$$ 1/4 is for the probability of one correct answer followed by the other 9 wrong answer probability multiplied by the number of ways this could happen.

I'm kinda confused on why we are multiplying by 10. Can someone explain? The exercise builds up to the definition of binomial distribution.

Answer 1

The probability that q. no. 1 (and only it) was answered correctly is

$$ {1 \over 4} \left({3 \over 4}\right)^9. $$

The probability that q. no. 2 (and only it) was answered correctly is

$$ {3 \over 4} {1 \over 4} \left({3 \over 4}\right)^8 = {1 \over 4} \left({3 \over 4}\right)^9. $$

The probability that q. no. 3 (and only it) was answered correctly is

$$ \left({3 \over 4}\right)^2 {1 \over 4} \left({3 \over 4}\right)^7 = {1 \over 4} \left({3 \over 4}\right)^9. $$

And so on.

There are 10 such possibilities. Since these events are disjoint, you need to add up the probabilities. This gives

$$ 10 {1 \over 4} \left({3 \over 4}\right)^9. $$

Answer 2

If there are 10 questions, and if you have answered only one correct (and 9 incorrectly), then there are 10 "slots" where the $\frac{1}{4}$ could be.