MHB What Is the Probability of Guessing Correctly on a Multiple Choice Test?

[Amad27]

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 is: $$\binom{10}{6} (0.2)^6 (0.8)^4$$

I see that, there $\binom{10}{6}$ ways of selecting $6$ correct questions here.

But then, take a question like:At a soccer match there are 230 all-stars and 220 half-stars. You pick five people from the crowd. What is the probability that exactly two are all-stars?

I would say:

$$P = \frac{\binom{230}{2} \cdot \binom{220}{3}}{\binom{450}{5}}$$

My question is, arent there also many ways to choose the $2$ all stars out of the $230$? So wouldn't you multiply by $\binom{230}{2}$ again?

 

[I like Serena]

My question is, arent there also many ways to choose the $2$ all stars out of the $230$? So wouldn't you multiply by $\binom{230}{2}$ again?

Hi Olok,

I'm not sure I understand your question.
$\binom{230}{2}$ is the number of ways to choose the $2$ all stars out of the $230$, which is a pretty large number.
What is the reason you think we should multiply by it twice?

Btw, your second formula represents the hypergeometric distribution instead of the binomial distribution. (Nerd)

 

[Amad27]

Hi Olok,

I'm not sure I understand your question.
$\binom{230}{2}$ is the number of ways to choose the $2$ all stars out of the $230$, which is a pretty large number.
What is the reason you think we should multiply by it twice?

Btw, your second formula represents the hypergeometric distribution instead of the binomial distribution. (Nerd)

No, because in the first one, we multiply by $\binom{n}{k}$ because the order matters, doesn't the order also matter in the second one?

I think I am messing this all up.

Are you using these definitions to solve probability like this? Like the definition of hypergeometric distribution, binomial distribution? Is it by definition that we use that?

 

[I like Serena]

No, because in the first one, we multiply by $\binom{n}{k}$ because the order matters, doesn't the order also matter in the second one?

I think I am messing this all up.

Whether order matters or not can be a bit confusing.
Actually, in both of these cases, we say that order does not matter.

For instance in the first case, it doesn't matter which of the questions are right, only how many of them.
Now suppose we want to know the probability that exactly the first 6 questions are right and the last 4 questions are wrong.
Then the order does matter and the probability is $0.2^6 \cdot 0.8^4$.
But since the order does not matter, the probability goes up by $\binom{10}{6}$, since that is the number of different configurations to have 6 questions right.

Are you using these definitions to solve probability like this? Like the definition of hypergeometric distribution, binomial distribution? Is it by definition that we use that?

The relevant definition, if we want to call it that, is that
$$P(\text{Desirable outcome}) = \frac{\text{Number of desirable outcomes}}{\text{Total number of outcomes}}$$
(Fine print: that is under the assumption that all outcomes are equally likely.)

The formulas are derived from that. Once you have them, it's easiest to use them rather than derive them again.