Suppose a student in Business Statistics has no been attending class this quarter but decides to take the exam anyway. If he randomly guesses on each of the 20 questions, then he has a 1 out of 5 chance of getting a correct answer, since it is multiple choice exam with choices a, b, c, d, or e. 8) How many questions should the student expect to get correct on this exam? 9) What is the probability that the student will get a “C” or better, which is 14 or more correct? 10) What is the probability that the student will score lower than a “C” (13 or less correct)?
+33654 The probability of getting an answer right is 1/5, so the expected number of right answers is (1/5)*20 = 4.
The probability of getting exactly 14 right is (1/5)14*(4/5)6*nCr(20,14) (nCr(20,14) is the numb er of ways of choosing 14 out of 20).
The probability of getting exactly 15 right is (1/5)15*(4/5)5*nCr(20,15)
The probability of getting exactly 16 right is (1/5)16*(4/5)4*nCr(20,16)
... etc. to
The probability of getting exactly 20 right is (1/5)20*(4/5)0*nCr(20,16)
Calculating all these and adding them up we find the probability of getting a C is approximately 2*10-6, so the probability of getting less than a C is 1 - 2*10-6.
+33654 The probability of getting an answer right is 1/5, so the expected number of right answers is (1/5)*20 = 4.
The probability of getting exactly 14 right is (1/5)14*(4/5)6*nCr(20,14) (nCr(20,14) is the numb er of ways of choosing 14 out of 20).
The probability of getting exactly 15 right is (1/5)15*(4/5)5*nCr(20,15)
The probability of getting exactly 16 right is (1/5)16*(4/5)4*nCr(20,16)
... etc. to
The probability of getting exactly 20 right is (1/5)20*(4/5)0*nCr(20,16)
Calculating all these and adding them up we find the probability of getting a C is approximately 2*10-6, so the probability of getting less than a C is 1 - 2*10-6.
