6. Chris is a very strong student who has an 85% chance of getting an A in any c

Question

6. Chris is a very strong student who has an 85% chance of getting an A in any course.

(a) Suppose that Chris takes one course at a time, and decides to keep taking courses until receiving the first non-A, and will quit after the first non-A. In any case, Chris will not take more than 4 courses. What is the expected number of courses Chris will take?

(b) What would the question have to be, in order to be able to use the N p shortcut to solve it, that is, for the answer to be N p = (4 courses)(.85 probability of an A in any given course) = 3.4 courses?

Explanation / Answer

a) Here the probability for the various values of N are computed as:

Therefore now the expected number of courses that Chris will take is computed as:

E(X)= 1*0.15 + 2*0.1275 + 3*0.108375 + 4*0.614125 = 3.186625

Therefore 3.186625 is the expected number of courses that would be taken by Chris

b) For using the formula of expected number as n*p = 0.85*4 = 3.4. This is actually the number of expected courses where out of the 4 courses taken Chris will score an A. Therefore this is a case of a binomial distribution that is the number of A grade courses in 4 courses.

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