anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
72 percent ? >,<
Directrix
  • Directrix
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. ------------------ 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.
Directrix
  • Directrix
Check out the binomial probability calculator at the link below: http://stattrek.com/online-calculator/binomial.aspx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I agree 72 percent
anonymous
  • anonymous
1 - ( P(0 correct) + P(1 correct)+P(2 correct) )\[1-\left(\left(\frac{3}{4}\right)^6+\left(\frac{3}{4}\right)^5\left(\frac{1}{4}\right)^1\frac{6!}{5! 1!}+\left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)^2\frac{6!}{4! 2!}\right)=\frac{347}{2048} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.