Question

A student who is trying to write a paper for a course has a choice of two topics, A and B. If topic A is chosen, the student will order two books through interlibrary loan, whereas if topic B is chosen, the student will order four books. The student believes that a good paper necessitates receiving and using at least half the books ordered for either topic chosen. If the probability that a book ordered through interlibrary loan actually arrives in time is 0.8 and books arrive independently of one another, which topic should the student choose to maximize the probability of writing a good paper? (Enter your answers to four decimal places.)

For topic A, P(at least half received)	=  .
For topic B, P(at least half received)	=  .

Which topic should the student choose if the arrival probability is only 0.6 instead of 0.8? (Enter your answers to four decimal places.)

For topic A, P(at least half received)	=  .
For topic B, P(at least half received)	=  .

Answer 1

Answer:

a)

Given,

To determine the required probabilities for topic A & B

P(arrival) = 0.8

Now for topic A:

P(X >= 1) = P(X = 1) + P(X = 2)

= 2C1*(0.8)^1*0.2^(2-1) + 2C2*0.8^2*0.2^(2-2)

= 2*0.16 + 1*0.64

= 0.32+0.64

P(X >= 1) = 0.96

For topic B:

P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4)

= 1 - [P(X = 0) + P(X = 1)]

= 1 - [4C0*0.8^0*0.2^4 + 4C1*0.8^1*0.2^(4-1)]

= 1 - [0.0016 + 0.0256]

= 1 - 0.0272

P(X >= 2) = 0.9728

Hence the topic B should be chosen since the probability of getting half the books ordered is higher than that of A

b)

Now p = 0.6

For topic A : P(X >= 1) = P(X = 1) + P(X = 2)

= 2C1*0.6^1*0.4^(2-1) + 2C2*0.6^2*0.4^2-2

= 0.48 + 0.36

P(X >= 1) = 0.84

Now for topic B :

P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4)

= 1 - [P(X = 0) + P(X = 1)]

= 1 - [4C0*0.6^0*0.4^(4-0) + 4C1*0.6^1*0.4^(4-1)]

= 1 - [0.0256 + 0.1536]

= 1 - 0.1792

P(X >= 2) = 0.8208

Hence the topic A should be chosen since the probability of getting half the books ordered is higher than that of B