# Suppose that you take a 20 question multiple-choice quiz by guessing. Each question has exactly one correct answer of the four alternatives given. What is the probability of getting all 20 questions correctly?

${\left(\frac{1}{4}\right)}^{20} = {1}^{20} / {4}^{20} \cong \frac{1}{1.1 \times {10}^{12}} = 1.1 \times {10}^{-} 12$

We can therefore say, very roughly, that the probability is slightly over one trillionth.

#### Explanation:

For each question, there is a probability of $\frac{1}{4}$ of guessing the right answer.

One the first question, therefore, we have the probability of $\frac{1}{4}$. If we then guess on the second question, we have another $\frac{1}{4}$ chance of getting that right. To see the odds of getting both right, we multiply the two probabilities, and so that's

$\frac{1}{4} \times \frac{1}{4} = {\left(\frac{1}{4}\right)}^{2} = \frac{1}{16}$

We can generalize and say that for any number of questions where we are guessing among 4 answers each time, the probability of getting them all right is:

${\left(\frac{1}{4}\right)}^{n}$, where $n$ is the number of questions.

For 20 questions, we have:

${\left(\frac{1}{4}\right)}^{20} = {1}^{20} / {4}^{20} \cong \frac{1}{1.1 \times {10}^{12}} = 1.1 \times {10}^{-} 12$

We can therefore say, very roughly, that the probability is slightly over one trillionth.