Question

A large industrial firm uses three local motels to provide overnight accommodations for its clients. From past experience it is known that 20% of the clients are assigned rooms at the Ramada Inn, 50% at the Sheraton, and 30% at the Lakeview Motor Lodge. If the plumbing is faulty in 5% of the rooms at the Ra- mada 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

(a) The probability is 0.054

(b) The probability is 0.4444

Explanation:

The probability P that a client will be assigned a room with faulty plumbing can be calculate as the sum of three probabilities: P1, P2 and P3

P1 is the probability that a client is assigned in Ramada Inn and the plumbing is faulty. This is calculate as:

P1 = 20%*5% = 0.01

P2 is The probability that a client is assigned in Sheraton and the plumbing is faulty. This is calculate as:

P2 = 50%*4% = 0.02

P3 is The probability that a client is assigned in Lakeview Motor Lodge and the plumbing is faulty. This is calculate as:

P3 = 30%*8% = 0.024

Then, the probability P is:

P = 0.01 + 0.02 + 0.024 = 0.054

For part b, the probability that a person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge is calculate as a division between P3 and P.

Where P3 is the probability that a person was assigned to Lakeview Motor Lodge and has a room with faulty plumbing and P is the probability that a person is going to be assigned a room with faulty plumbing.

So, the probability that a person with a room having faulty plumbing was assigned accommodations at the Lakeview Motor Lodge is:

Probability = 0.024/0.054 = 0.4444