Probability Distribution Exercises

#1
Hi guys,

I got some big problems in Statistics and have actually no clue about 3 exercises I have to solve. If anyone can help me (even if its just one exercise) I would be so grateful. I will post them below:

Question 1
A student wants to apply for a job offered by an attractive and very wellknown
company. The student expects to have very good exam results.
The probability to be invited for an interview and to be employed by the
firm is 5 per cent. This disencourages the student. Later the student hears
that the probability of having very good exam results given that the
company employs an applicant is 99 per cent. Furthermore he obtains the
information that the probability of having very good exam results given
that the company does not employ an applicant is 5 per cent. What is the
probability for the student to be employed given that he has very good
exam results?

Question 2

In a hotel it has been estimated that the average number of guests
staying overnight is 100. It has also been found out that every day 20
guests leave the hotel. If we randomly ask 10 guests in the morning how
large is the probability that exactly 5 of them will leave the hotel?

Question 3

A tourist shop sells snacks including peanuts. When the stock of peanuts
drops to 25 tins, a replenishment order is placed. It has been determined
that the demand for peanuts during the replenishment lead-time is
normally distributed with a mean of 20 tins and a standard deviation of 10
tins. What is the probability that the demand during replenishment is
greater than 25 tins?


Thank you guys in advance.
 
#2
Q1 is about conditional probability. You’ll need the following identities:
P(A|B) × P(B) = P(B|A) × P(A) [Bayes’ theorem]
P(¬X) = 1 – P(X)
P(¬X|Y) = 1 – P(X|Y)​
With A = “Interviewed & employed” and B = “Very good exam results”, you can set up the knowns and unknowns and their relationships with those identities, and using a little algebraic manipulation you can solve for P(A|B) and P(B).

Q2 is about a Poisson distribution with parameter λ = E(X) = 2 since we expect to have 2 guests leaving from 10 randomly selected ones.

Q3 is about a normal distribution with μ = 20 and σ = 10. For x = 25, z = (25 – 20)/10 = 0.5 and the required probability P(x > 25) is the area under the N(0,1) curve for z > 0.5.