Imagine taking a multiple choice test that has three possible answers to each question. This means that even if you don’t know any answer, your chance of getting a question right is still 1/3. How likely is it to get all questions right by guessing if the test contains ten questions?

Here we are looking at the event “correct answer” which occurs with a probability of p(correct answer) = 1/3. We want to know the odds of this event happening ten times in a row. For that we simply apply the multiplication rule:

- p(all correct) = (1/3)
^{10} = 0.000017

Doing the inverse, we can see that this corresponds to about 1 in 60000. So if we gave this test to 60000 students who only guessed the answers, we could expect only one to be that lucky. What about the other extreme? How likely is it to get none of the ten questions right when guessing?

Now we must focus on the event “incorrect answer” which has the probability p(incorrect answer) = 2/3. The odds for this to occur ten times in a row is:

- p(all incorrect) = (2/3)
^{10} = 0.017

In other words: 1 in 60. Among the 60000 guessing students, this outcome can be expected to appear 1000 times. How would these numbers change if we only had eight instead of ten questions? Or if we had four options per question instead of three? I leave this calculation up to you.

### Like this:

Like Loading...

*Related*