Question

A quiz consists of 6 multiple choice questions.each question has 4 possible answers.a student is unprepared and guess answers by random.he passes if he gets atleast 3 questions right.what is the probability that he will pass

Answer 1

72 percent ? >,<

Answer 2

Assuming that the student writes an answer to each of the six questions.
The probability of getting a single question correct by guessing is 1/4.
The probabiliy of getting a single question wrong by guessing is 3/4.
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P(3 correct) = C(6,3 )* (1/4) ^ 3 * (3/4) ^ 3 = 135 /1024
P (4 correct) = C(6,4) * (1/4) ^ 4 * (3/4) ^ 2 = 135 / 4096
P(5 corrrect) = C(6,5) * (1/4) ^ 5 * (3/4) ^ 1 = 9 / 2048
P(6 correct) = C(6,6) * (1/4) ^ 6 * (3/4) ^ 0 = 1 / 4096
Sum of probabilities above is 347 / 2048 = .1694 approx or .17.

If you take the test 100 different times, expect to pass 17 times.

Answer 3

Check out the binomial probability calculator at the link below:

http://stattrek.com/online-calculator/binomial.aspx

Answer 4

I agree 72 percent

Answer 5

1 - ( P(0 correct) + P(1 correct)+P(2 correct) )\[1-\left(\left(\frac{3}{4}\right)^6+\left(\frac{3}{4}\right)^5\left(\frac{1}{4}\right)^1\frac{6!}{5! 1!}+\left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)^2\frac{6!}{4! 2!}\right)=\frac{347}{2048} \]