# math-Probablity

A student is writing a multiple choice test consisting of 40 questions, each of which provides four possible choices. He is certain that he has 16 questions correct. If he guesses for all of the remaining 24 questions, what is the probability that he will pass the test?

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1. Assuming that a pass is getting at least 20 of the 40 correct ....
What the student CANNOT have is
zero of the remaining 24 correct : C(24,0) (1/4)^0 (3/4)^24 = .001003391
1 of the remaining 24 correct : C(24,1) (1/4) (3/4)^23 = .00802713
2 of the remaining 24 correct : C(24,2) (1/4)^2 (3/4)^22 = .030770665
3 of the remaining 24 correct : C(24,3) (1/4)^3 (3/4)^21 = .075217183

prob he will pass = 1 - (sum of the above 4 cases)
= ...

check my arithmetic

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2. Assuming:
1. he got indeed the 16 questions correctly.
2. the remaining 24 answers were random choices
3. there was exactly one correct answer
4. passing grade is 50%

to each of the remaining questions
then he will need 4 more correct answers to get 50%.
Let
p=probability of guessing a question correctly = 0.25
q=1-p=probability of guessing incorrectly = 0.75

The probability of getting 0 to 3 correct guesses (i.e. fail) is
C(24,0)*q^24*p*0 +
C(24,1)*q^23*p*1 +
C(24,2)*q^22*p*2 +
C(24,3)*q^21*p*3
=0.0010+0.0080+0.00308+0.00752
=0.1150
where C(n,r)=n!/((n-r)!r!)
Therefore the probability that he will pass (at 50%) is
1-0.1150=0.8850

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3. Here we go again!

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4. What is the probability of that ? lol

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