SOLUTION: A) A test consists of 10 true and false questions. To pass the test a student must answer at least eight questions correctly. If the student guesses on each questions what is the

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Question 296701: A) A test consists of 10 true and false questions. To pass the test a student must answer at least eight questions correctly. If the student guesses on each questions what is the probability that the student will pass the test?
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A test consists of 10 true and false questions. To pass the test a student must answer at least eight questions correctly. If the student guesses on each questions what is the probability that the student will pass the test?
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Binomial Problem with n = 10 and p = 1/2
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P(8<= x <= 10) = 1 - P(0<= x <=7) = 1 - binomcdf(10,0.5,7)
= 0.0547..
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Cheers,
Stan H.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
Assume it was only 3 answers on the test and the student needed to get at least 2 out of 3 correct.

Probability of getting exactly 0 wrong would be .5^3 = .125 * 1 = .125
Probability of getting exactly 1 wrong would be .5^3 = .125 * 3 = .375
Probability of getting exactly 2 wrong would be .5^3 = .125 * 3 = .375
Probability of getting exactly 3 wrong would be .5^3 = .125 * 1 = .125

Total probability is equal to 1 as it should be.

Probability of getting 0 or 1 wrong would be .375 + .125 = .5

Since 0 or 1 wrong is the same as getting 2 or 3 right, then this is the probability that the student will get at least 2 right.

The individual probabilities are multiplied by the number of ways they can occur.

If we let 0 = wrong and 1 = correct, then:

You can get 0 wrong only 1 way (111)
You can get 1 wrong 3 ways (110) (101) (011)
You can get 2 wrong 3 ways (001) (010) (100)
You can get 3 wrong 1 way (000)

The same concept applies to the larger numbers.

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With 10 answers, this is what happens:

p(0) = probability of getting exactly 0 correct.
p(1) = probability of getting exactly 1 correct.
etc.

p(0) = .5^10 = .000976563 * 1 = .000976563
p(1) = .5^10 = .000976563 * 10 = .009765625
p(2) = .5^10 = .000976563 * 45 = .043945313
p(3) = .5^10 = .000976563 * 120 = .1171875
p(4) = .5^10 = .000976563 * 210 = .205078125
p(5) = .5^10 = .000976563 * 252 = .24609375
p(6) = .5^10 = .000976563 * 210 = .205078125
p(7) = .5^10 = .000976563 * 120 = .1171875
p(8) = .5^10 = .000976563 * 45 = .043945313
p(9) = .5^10 = .000976563 * 10 = .009765625
p(10) = .5^10 = .000976563 * 1 = .000976563

Total probability equals 1 as it should.

Probability of getting 0 or 1 or 2 wrong is equal to:

.000976563 + .009765625 + .043945313 = .0546875

Since the probability of getting 0 or 1 or 2 wrong is the same as the probability of getting 8 or 9 or 10 right, then the probability that the student will get at least 8 correct is equal to .0546875.

The number of ways each percentage can be achieved is given by the formula:

%28n%21%29+%2F+%28%28x%21%29%2A%28n-x%21%29%29

For example, with 10 answers, the number of ways of getting exactly 4 wrong is equal to:



With 10 answers, the number of ways of getting exactly 6 wrong is the same as getting exactly 4 wrong as shown below: