Standard multiple choice probability

A student is presented with a test consisting of standard multiple choice problems: each problem has four options, and they may select only one.

The student is guessing randomly on each question. Skill is not a factor.

What is the probability that the student will get **at least** a passing score of 70% on the test?

Re: Standard multiple choice probability

I think you should say p=0.25 and n= the total number of questions, do you know how many there is?

Then you need to find x s.t.

$\displaystyle \displaystyle P(x) = \binom{n}{x} 0.25^x\times 0.75^{n-x} \geq 0.7$

Re: Standard multiple choice probability

Quote:

Originally Posted by

**pickslides** I think you should say p=0.25 and n= the total number of questions, do you know how many there is?

Then you need to find x s.t.

$\displaystyle \displaystyle P(x) = \binom{n}{x} 0.25^x\times 0.75^{n-x} \geq 0.7$

I appreciate your help, but I also have to confess that this is a bit beyond me. You've told me *what* to do, but I don't know *how* to do it. That is, I don't know how to actually execute that formula. I have no problem with exponents, etc. It's mainly this part, which I don't recognize at all:

$\displaystyle \binom{n}{x}$

Also, you said p=0.25 and n=the total number of questions. What is x?

Now, I'm totally willing to do the dirty work if I know where to look. I now know what I'm looking for, many thanks to you. Any chance you could tell me where to find a resource that would actually break down how to perform this calculation?

Re: Standard multiple choice probability

Quote:

Originally Posted by

**jbwtucker**

$\displaystyle \binom{n}{x}$

$\displaystyle \displaystyle \binom{n}{x} = \frac{n!}{x!(n-x)!}$

where $\displaystyle n! = n\times (n-1) \times (n-2)\times \dots \times 3\times 2\times 1$

Does this make sense?

Quote:

Originally Posted by

**jbwtucker**

Also, you said p=0.25 and n=the total number of questions. What is x?

$\displaystyle x$ is the number of questions the student needs to answer correctly to get 70% on the test.

$\displaystyle n$ is the total number of questions on the test. Do you know how many questions make up the test? This is important if we are trying to work out how many the student needs to answer correctly.

Re: Standard multiple choice probability

Thank you, this is awesome! It does indeed make sense, and I can do it.

Thank you also for clarifying re: x and n. The total number of questions is not yet set, but it will be, and the important thing is that now that I understand how to execute this formula, I can use it to find the answer I need, even if the variables change.

Thanks so much!