Need detail solution!!!!!!!! Problem 3: A graduating MIE student goes to a caree

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Need detail solution!!!!!!!!

Problem 3: A graduating MIE student goes to a career fair, and his/her likelihood of receiving an interview invitation at a booth visit depends on how well he/she did in MIE263 (interviews will be immediately scheduled for a later date). Specifically, an A in MIE263 results in a probability p 0.99 of obtaining an interview invitation, whereas a C in MIE263 results in a probability p - 0.12 of an invitation. Assume the probability of each student getting an interview is independent of the other students. a) Find the PMF for the random variable Y that denotes the number of career fair booth visits a student must make before his/her first invitation (including the visit that results in invitation). Express your answer in terms ofp b) On average, how many booth visits must an A student make before getting an interview? How about a C student? c) Assume that an A student and two C students go to the career fair. If each student can go to at most 5 booths, what is the probability that the A student doesn't get an interview and at least one of the C student gets at least one interview? d) If the A student decides to go to booths until he/she receives 3 interview invitations, Write down the PMF of the number of booths that he/she has to visit. How many booths does he/she have to visit on average? How about a C student? e) Assume that the A student and the C student both visit random booths and get together after each booth visit to share their result. They stop visiting booths when on one round of visits they both get an interview. Let R be the total number of booth visits per student. What is the PMF of R? What is the expected value of R?

Explanation / Answer

a) f(Y) = (1-p)^(y-1)*p

b) E(y) = 1/p

For A students, E(y) = 1/.99 = 100/99 = 1 1/99

For C students, E(y) = 1/.12 = 100/12 = 25/3 = 8 1/3

c) P(A student does not get an interview) = P(5 failures) = (1-.99)^5

P(1 of 2 C students gets at least 1 interview) = 1 - P(10 failures) = 1 - (1-.12)^10 = 1 - .88^10

P(A student doesn't get an interview and C student does) is

(1-.99)^5(1 - .88^10) = (.01)^5 - (.01)^5(.88)^10 = 7.21499023990598E-11

d) This is the negative binomial This may also be figured as the probability of 2 successes in n-1 interviews and a success in the n'th interview. It is

f(n) = C(n-1,2)(1-p)^(n-3)p^3 As p = .99, this is C(n-1,2).01^(n-3).99^3

The number of booths visited by the A student on average is 3/p = 3/.99 = 100/33 = 3 1/33

For C students, the average is 3/.12 = 25

e) P both are successful in a given round is .99*.12 = .1188

f(R) = .1188(1-.1188)^(R-1) = .1188*.8812^(R-1)

f) P(at least 1 gets an interview) = 1 - P(both fail) = 1 - (1-.12)(1-.99) = 1 - .88*.01 = .9912

f(N) = .9912(1-.9912)^(N-1) = .9912*.0088^(N-1)

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