Posted by demi on Tuesday, June 5, 2012 at 3:16pm.
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
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
Here we go again!
What is the probability of that ? lol
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