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November 26, 2014

November 26, 2014

Posted by **demi** on Tuesday, June 5, 2012 at 3:16pm.

- math-Probablity -
**Reiny**, Tuesday, June 5, 2012 at 3:41pmAssuming 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

- math-Probablity -
**MathMate**, Tuesday, June 5, 2012 at 3:41pmAssuming:

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

- To Reiny: great minds... -
**MathMate**, Tuesday, June 5, 2012 at 3:45pmHere we go again!

- math-Probablity -
**Reiny**, Tuesday, June 5, 2012 at 3:51pmWhat is the probability of that ? lol

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