Question 1206840: A test consists of 10 multiple choice questions, each with five possible answers, one of which is correct. To pass the test a student must get 60% or better on the test. If a student randomly guesses, what is the probability that the student will pass the test?
A. 0.759
B. 0.377
C. 0.006
D. 0.060
Answer by math_tutor2020(3817)   (Show Source):
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Answer: 0.006 (choice C)
Work Shown
n = 10 questions
p = 1/5 = 0.2 = probability of getting the correct answer on any given question
x = number of correct questions
x is a whole number and spans from x = 0 to x = 10
60% of 10 = 0.60*10 = 6
if x = 6, 7, 8, 9 or 10, then the student passes the exam.
We have a binomial distribution since...- Each question effectively boils down to two options (either the student gets the correct answer or they don't).
- Each trial has the same probability.
- Each trial is independent of one another.
The binomial probability formula is
B(x) = (nCx)*(p)^x*(1-p)^(n-x)
where the nCx refers to the nCr combination formula.
To compute the nCr value in a spreadsheet, use the command called Combin
For example, type in =Combin(10,7) to mean 10C7.
Do not use the "choose" command. That's something entirely different.
To see how to compute nCr values on a TI84, check out this article
Let's plug in x = 6
B(x) = (nCx)*(p)^x*(1-p)^(n-x)
B(6) = (10C6)*(0.2)^6*(1-0.2)^(10-6)
B(6) = (210)*(0.2)^6*(1-0.2)^(10-6)
B(6) = 0.005505024
This is the probability of getting exactly 6 questions correct.
Repeat similar steps to find these other values
B(7) = 0.000786432
B(8) = 0.000073728
B(9) = 0.000004096
B(10) = 0.0000001024
Then add them up to get
0.005505024 + 0.000786432 + 0.000073728 + 0.000004096 + 0.0000001024 = 0.0063693824
That then rounds to 0.006 which is the final answer.
This is the probability of getting 6 or more questions correct.
On a TI84 we can find this answer quickly by computing 1 - BinomCDF(10,0.2,5)
The BinomCDF(10,0.2,5) portion adds up the binomial probability values from x = 0 to x = 5.
Subtract that from 1 to get the sum from x = 6 to x = 10.
In other words,
B(0)+B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)+B(8)+B(9)+B(10) = 1
which rearranges to
B(6)+B(7)+B(8)+B(9)+B(10) = 1 - ( B(0)+B(1)+B(2)+B(3)+B(4)+B(5) )
The stuff in blue is the BinomCDF(10,0.2,5) portion.
Review this article for more info.
On a spreadsheet the command would be =1-BinomDist(5,10,0.2,1)
The input arguments are in a slightly different order compared to BinomCDF, but it's the same basic idea. The '1' at the end lets the spreadsheet know you want a CDF instead of a PDF.
Don't forget about the equal sign up front.
Review this article for more info.
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