You have a 5-question multiple-choice test. Each question has four choices. You don’t know any of the answers. What is the experimental probability that you will guess exactly three out of five questions correctly?
I would use the binomial distribution method
ok, so, i know how to do that but, which numbers do i pull from the text?
We would use (n x) p^x q^(n-x)
where p=.25 q=.75 n=5 x=3
woah... I don't think i've ever used that before but I'll try it! ^^
by (n x) i mean n choose x
I got I got about 8.79%
Do you have the solution?
im not really sure how to get that, so (5x3) (2.5)^3 75 ^(5-3) ?
5C3 not 5x3
ooh, ok i got it now ^^ 5C3 is 10 right? thats what i get when i plug it in
yes
oh and p=.25 and q=.75 not 2.5 and 75
I can explain the binomial distribution to you if you would like
i get the foiling and unfoiling method, is it the same thing? I don't think i'v ever done it with exponents before
I am not sure what the foiling unfoiling method is.
For this problem I believe most people would do what I did. Although there are other ways to think about it.
would you mind explaining the binomial distribution, if its not to much trouble?
Not at all. I had a horrible professor for this stuff and taught myself... it took me forever!!
Ok so I like to think of the binomial distribution as basically a little shortcut we can use to find the probability of something occurring. It has many restrictions though and can only be used under certain circumstances.
First of all: Each trial (or question) must have an outcome of either yes or no... in this case it was right or wrong...
It must consist of repeated trials that do not depend on each other... (i.e. getting the first question correct has no influence on the next question.)
And the probability of "yes" (or getting the question correct) must remain the same.
If the scenario satisfies these three things then we are allowed to use the binomial distribution.
You following thus far?
yes, so far, yes.
sweet!
Here is the formula we use to calculate probability using the binomial distribution: \[_n C_{x}* p^x *q^{n-x}\]
n is the number of trials (questions) x is the number of success we want (# of correct answers) p is the probability of getting a success on any one of the trials q is the probability of getting a failure on any one of the trials
when you plug all those numbers in the formula I posted will give you the probability of getting x successes
Does it make more sense now?
ok yes theoretically it makes sense in my head but when i go to plug it in, 5C3 * p^ (what is the exponent here?) * .75^(5-3)
x
which in this case is 3
we want 3 correct answers
p is .25.. you have a 1 in 4 chance of getting the answer correct on each trial
Get it?
sorry my computer died, yes!! I understand! thank you sososo much really!
Join our real-time social learning platform and learn together with your friends!