Question 1023341: A multiple choice exam has 12 questions and each question has options a-e to choose from. There is only one correct answer for each question. Assume a student is brought in that has never taken the class, and is just guessing random answers.
a.) What is the probability he gets them all correct?
I used the binomial formula: (12!)/(12!-12!) x (.2)^12 x (.8)^0 = 4.096 x 10^-9
I dont understand how to make that answer into a percentage/did I do the formula correct for this problem?
b.) What is the probability that he gets more than 2 correct?
Do I need to use the binomial formula starting with 3 as k and then do every number as k up to 12? That seems like too much work to do the binomial formula 9 times and then add all those answers up. Is there an easier way to solve for the probability of getting more than 2 correct?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Just like converting any other number to a percentage; you move the decimal point two places to the right. So:
If you want to calculate the probability of "more than 2 correct" straight up, then you are right, you have to calculate the probability of exactly 3 plus the probability of exactly 4, plus...and so on all the way up to 12. However, the probability of MORE THAN 2 is equal to 1 minus the probability of 2 OR LESS. Calculate the probability of 0, 1, and 2, add the three probabilities, and subtract from 1.
\ =\ \sum_{k\ =\ 3}^{12}{{12}\choose{k}}\left(0.2\right)^k\left(0.8\right)^{12\,-\,k}\ =\ 1\ -\ P_{12}(< 3,0.2)\ =\ 1\ -\ \sum_{k\ =\ 0}^{2}{{12}\choose{k}}\left(0.2\right)^k\left(0.8\right)^{12\,-\,k})
If you have either Excel (on a Windows machine) or Numbers (on a Mac), open a spreadsheet and enter
)
John
My calculator said it, I believe it, that settles it
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