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

In summary, the conversation discusses the probability of getting exactly 6 questions correct on a test with 10 multiple choice questions and 5 choices for each question, if one guesses on all answers without reading the questions. The first formula presented is for the binomial distribution and the second is for the hypergeometric distribution. The conversation also touches on the concept of order not mattering in these calculations. The relevant definition for calculating probabilities is mentioned as well.
  • #1
Amad27
412
1
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?
 
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  • #2
Olok said:
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)
 
  • #3
I like Serena said:
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?
 
  • #4
Olok said:
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.
 

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

What is the Binomial Distribution Issue?

The Binomial Distribution Issue is a statistical concept that deals with the probability of obtaining a certain number of successes in a fixed number of trials, where each trial has two possible outcomes - success or failure. It is often used to analyze data in fields such as biology, psychology, and economics.

What are the key components of the Binomial Distribution?

The key components of the Binomial Distribution include the number of trials (n), the probability of success (p), and the number of successes (x). These components are used to calculate the probability of obtaining a specific number of successes in a given number of trials.

How is the Binomial Distribution different from other probability distributions?

The Binomial Distribution is different from other probability distributions in that it only considers two possible outcomes for each trial, whereas other distributions may have multiple outcomes. Additionally, the Binomial Distribution assumes that each trial is independent and that the probability of success remains constant throughout all trials.

How can the Binomial Distribution be applied in real-world situations?

The Binomial Distribution can be applied in a variety of real-world situations, such as predicting the likelihood of a certain number of people purchasing a product, the probability of a medical treatment being successful, or the chances of winning a game with a fixed number of attempts.

What are some limitations of the Binomial Distribution?

One limitation of the Binomial Distribution is that it assumes each trial is independent, which may not always be the case in real-world scenarios. It also requires a fixed number of trials, which may not always be applicable. Additionally, it is most accurate when the probability of success is close to 0.5, so it may not be suitable for situations with extreme probabilities.

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