SOLUTION: a multiple choice test contain 25 questions each with four answers. assume that a student just guesses on each question. (a)what is the probability that the student answers more t

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Question 1133310: a multiple choice test contain 25 questions each with four answers. assume that a student just guesses on each question. (a)what is the probability that the student answers more than 20 question correctly? (b) What is the probability that the student answer fewer than five questions correctly
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

CORRECTED answer below -- search for CORRECTED
The probability of getting exactly n correct is:
+P%28n_corr%29+=+%281%2F4%29%5E%28n%29%2A%283%2F4%29%5E%2825-n%29+

So we can build this table:

perl -e 'for($c=0; $c<26; $c++) { $p = (1/4)**($c)*(3/4)**(25-$c); printf "Pr(%d correct) = %e\n", $c, $p; }'
Pr(0 correct) = 7.525435e-04
Pr(1 correct) = 2.508478e-04
Pr(2 correct) = 8.361594e-05
Pr(3 correct) = 2.787198e-05
Pr(4 correct) = 9.290660e-06
Pr(5 correct) = 3.096887e-06
Pr(6 correct) = 1.032296e-06
Pr(7 correct) = 3.440985e-07
Pr(8 correct) = 1.146995e-07
Pr(9 correct) = 3.823317e-08
Pr(10 correct) = 1.274439e-08
Pr(11 correct) = 4.248130e-09
Pr(12 correct) = 1.416043e-09
Pr(13 correct) = 4.720144e-10
Pr(14 correct) = 1.573381e-10
Pr(15 correct) = 5.244605e-11
Pr(16 correct) = 1.748202e-11
Pr(17 correct) = 5.827339e-12
Pr(18 correct) = 1.942446e-12
Pr(19 correct) = 6.474821e-13
Pr(20 correct) = 2.158274e-13
Pr(21 correct) = 7.194245e-14
Pr(22 correct) = 2.398082e-14
Pr(23 correct) = 7.993606e-15
Pr(24 correct) = 2.664535e-15
Pr(25 correct) = 8.881784e-16


(a) Adding the last 5 lines gives:
P(more than 20 are correct) = +1.075x10%5E%28-13%29+

(b) Adding the first 5 lines gives:
P(less than 5 are correct) = +1.124x10%5E%28-3%29+

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Hmmm, I think I've made a mistake -- the sum of the table should be 1.000 (as one of the cases must certainly occur), but it is not. I will investigate and post the fix later...
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================= Ignore solution above, CORRECTED below ==============

Doh! P(exactly r correct) = +%281%2F4%29%5Er%2A%283%2F4%29%5E%2825-r%29%2AnCr+ where n=25, r=number correct. I will fix the table shortly...
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Corrected solution:
perl -e '$sum = 0; for($c=0; $c<26; $c++) { $ncr=1; for($i=25;$i>(25-$c);$i--) { $ncr *= $i; } for ($i=1; $i<=$c; $i++) { $ncr /= ($i);} $p = (1/4)**($c)*(3/4)**(25-$c)*$ncr; printf "Pr(%d correct) = %e\n", $c, $p; $sum+=$p; } print "sum = $sum\n"; '
Pr(0 correct) = 7.525435e-04
Pr(1 correct) = 6.271195e-03
Pr(2 correct) = 2.508478e-02
Pr(3 correct) = 6.410555e-02
Pr(4 correct) = 1.175268e-01
Pr(5 correct) = 1.645376e-01
Pr(6 correct) = 1.828195e-01
Pr(7 correct) = 1.654082e-01
Pr(8 correct) = 1.240561e-01
Pr(9 correct) = 7.810941e-02
Pr(10 correct) = 4.165835e-02
Pr(11 correct) = 1.893561e-02
Pr(12 correct) = 7.363850e-03
Pr(13 correct) = 2.454617e-03
Pr(14 correct) = 7.013190e-04
Pr(15 correct) = 1.714335e-04
Pr(16 correct) = 3.571532e-05
Pr(17 correct) = 6.302704e-06
Pr(18 correct) = 9.337339e-07
Pr(19 correct) = 1.146691e-07
Pr(20 correct) = 1.146691e-08
Pr(21 correct) = 9.100720e-10
Pr(22 correct) = 5.515588e-11
Pr(23 correct) = 2.398082e-12
Pr(24 correct) = 6.661338e-14
Pr(25 correct) = 8.881784e-16



(a) Adding the last 5 lines gives:
P(more than 20 are correct) = +9.677x10%5E%28-10%29+

(b) Adding the first 5 lines gives:
P(less than 5 are correct) = +2.137x10%5E%28-1%29+