SOLUTION: How would you go about solving this problem? I am having difficulty finding an example in my notes.
A multiple-choice test has 7 questions each with 5 responses, one of which is
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A multiple-choice test has 7 questions each with 5 responses, one of which is
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Question 668672: How would you go about solving this problem? I am having difficulty finding an example in my notes.
A multiple-choice test has 7 questions each with 5 responses, one of which is correct. The lowest number of correct answers a student must obtain in order to receive a grade of A is 5. Find the probability of receiving an A by randomly and independently guessing at the correct answer. Found 2 solutions by stanbon, swincher4391:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A multiple-choice test has 7 questions each with 5 responses, one of which is correct. The lowest number of correct answers a student must obtain in order to receive a grade of A is 5. Find the probability of receiving an A by randomly and independently guessing at the correct answer.
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Binomial Problem with n = 7 ; p(correct) = 1/5 ; p(wrong) = 4/5
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P(5 <= x <=7)
= 1 - binomcdf(7,1/5,4)
= 1 - 0.9953
= 0.0047
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Cheers,
Stan H.
You can put this solution on YOUR website! With each question having 1 right answer out of 5, p = 1/5 = .2
Now in order to get an A you must get 5 or more correct. So, basically we are trying to find the P[5 correct] + P[6 correct] + P[7 correct]
P[5 correct] = (7 choose 5) * (.2)^5 * (.8)^2 because we are choosing 5 that we want to get right out of 7, there's a .2 chance of getting each right (done 5 times) and there's a .8 chance of getting it wrong (done 2 times)
P[6 correct] = (7 choose 6) * (.2)^6 * (.8)^1
P[7 correct] = (7 choose 7) * (.2)^7 * (.8)^0
So P[>= 5 correct] = P[5 correct] + P[6 correct] + P[7 correct] = (7 choose 5) * (.2)^5 *(.8)^2 + (7 choose 6) * (.2)^6 * (.8)^1 + (.2)^7 =
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