Question 261826: I understand basic probability but am very confused when it comes to the below question. I don't even have a clue where to start to solve this question.
A multiple choice test has 20 questions with each having 4 possible answers with one correct. Assume a student randomly guesses the answer to every question.
a. What is the probability of getting exactly 11 correct answers?
20x4=80 1 right answer from 1 question=1/4
b. What is the probability of getting less than 9 correct answers?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A multiple choice test has 20 questions with each having 4 possible answers with one correct. Assume a student randomly guesses the answer to every question.
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This is a binomial problem because the probability of guessing the right
answer on each problem is 1/4 and there are n= 20 such problems.
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a. What is the probability of getting exactly 11 correct answers?
P(x+11) = 20C11(1/4)^11(3/4)9 = binompdf(20,1/4,11) = 0.0030
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b. What is the probability of getting less than 9 correct answers?
P(0<= x <=8) = binomcdf(20,1/4,8) = 0.95907...
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I used a TI calculator to get these answers.
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Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
You need the probability of  successes out of  trials where the probability of success on any given trial is  and the probability of failure on any given trial is 
\ =\ \left(n\cr k\right)p^kq^{n-k})
where ) is the number of ways to select things from things which is equal to !})
Your probability of success for a 4 choice test item is  . Consequently your probability of failure for a 4 choice test item is  .
Your is 20 because there are 20 questions and your is 11 because you want the probability of exactly 11 correct. So:
\ =\ \left(20\cr 11\right)\left(\frac{1}{4}\right)^{11}\left(\frac{3}{4}\right)^{9}\ =\ \frac{20!}{11!9!}\left(\frac{1}{4}\right)^{11}\left(\frac{3}{4}\right)^{9})
Looks like it is time to get out the calculator.
For part b. You need the probability of all possibilities of getting less than 9 correct. So, that is the probability of getting 0 correct plus the probability of getting 1 correct plus...and so on...plus the probability of getting 8 correct. Figure out each of those probabilities the same way as we did in part a and then sum them.
The actual formula for part b is:
_{cum}\ =\ \sum_{i=0}^k{\left(n\cr i\right)p^iq^{n-i}})
If you just want the answers quickly and you have access to MS Excel, then open a blank spreadsheet and, for part a, type the following into any blank cell:
 =BINOMDIST(11,20,0.25,FALSE)
And for part b
=BINOMDIST(8,20,0.25,TRUE)
John
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