Question

Each week, Stéphane needs to prepare 4 exercises for the following week's homework assignment. The number of problems he creates in a week follows a Poisson distribution with mean 6.9.

a. What is the probability that Stéphane manages to create enough exercises for the following week's homework? Round your answer to 4 decimal places.

b. Unfortunately, each week there is a 42% chance that a visiting scholar from Switzerland arrives and burdens Stéphane with research questions all week. During these weeks he only writes an average of 3.45 exercises. If Stéphane fails to write 4 exercises one week, what is the probably that he received a visiting scholar that week? Round your answer to 4 decimal places.

c. The last week of the semester, Stéphane decides to "reward" the students by no longer limiting himself to 4 exercises, and instead assigning every exercise he writes. If a student with a 55% chance of correctly answering an exercise is expected to answer 2.75 questions correctly, what is the probably that Stéphane did not have a visitor that week? Round your answer to 4 decimal places.
Hint: First find the number of exercises in the last week of the semester from the chance and expected value of the correct answers.

Answer 1

Part (a)

Required probability

= P(Stephane creates at least 4 exercises)

= P(X >= 4), where X follows Pois(6.9)

= 0.9129 (approximately)

Part (b)

Let us write A be the event that Stephane received a visiting scholar in the current week and B be the event that Stephane fails to create 4 exercises.

Then by the problem, P(A) = 0.42

Also, A^c is the event that Stephane did not receive a visiting scholar, with P(A^c) = 1 - 0.42 = 0.58

Required probability

Part (c)

Suppose Stephane created k exercises in the last week of the semester. Then the expected number of corrected questions by a student with 55% chance of correctly answering each question is 0.55 * k, assuming that the the probability of correct answer is the same for all the exercises.

By the problem, 0.55 * k = 2.75, which implies k = 5.

Let C be the event that Stephane created 5 exercises in the last week.

Then required probability