Question

Please answer the problem below for all parts. Please show all work and write clearly. Thanks.

(Apply total probability and Bayes’ rules)

A large industrial firm uses three local motels to provide overnight accommodations for its clients. From past experience it is known that 22% of the clients are assigned rooms at the Ramada Inn, 50% at the Sheraton, and 28% at the Lakeview Motor Lodge. If the plumbing is faulty in 5% of the rooms at the Ramada Inn, in 4% of the rooms at the Sheraton, and in 8% of the rooms at the Lakeview Motor Lodge, what is the probability that:

(a) A client will be assigned a room with faulty plumbing?

(b) A person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge?

Answer 1

Let

R: a client is assigned accommodation at Ramada Inn

Given, P(R) = 0.22

S: a client is assigned accommodation at Sheraton

Given, P(S) = 0.50

L: a client is assigned accommodation at Lakeview Motor Lodge

Given, P(L) = 0.28

P(F|R): Probability of faulty plumbing rooms in Ramada = 0.05

P(F|S): Probability of faulty plumbing rooms in Sheraton = 0.04

P(F|L): Probability of faulty plumbing rooms in Lakeview = 0.08

Answer a)

Now, we need to find probability that client is assigned with faulty plumbing room P(F)

Using theorem of total probability, we get:

P(F) = P(R)*P(F|R) + P(S)*P(F|S) + P(L)*P(F|L) = 0.22*0.05 + 0.50*0.04 + 0.28*0.08

P(F) = 0.0534

Answer b)

Now, we need to find probability that faulty plumbing room was assigned at Lakeview Motor Lodge P(L|F)

Using Bayes' Theorem, we get:

P(L|F) = [P(F|L)*P(L)]/[P(R)*P(F|R) + P(S)*P(F|S) + P(L)*P(F|L)]

P(L|F) = 0.28*0.08/(0.22*0.05 + 0.50*0.04 + 0.28*0.08)

P(L|F) = 0.28*0.08/0.0534

P(L|F) = 0.4195