Question 1192249
Here's how to solve this probability problem:

First, let's organize the data into a table:

| Hotel | Lodging Only | Lodging with Breakfast | Total |
|---|---|---|---|
| A | 30 | 80 (110 total - 30 lodging only) | 110 |
| B | 90 (210 total - 120 with breakfast) | 120 | 210 |
| Total | 120 | 200 | 300 |

**(a) Probability of Hotel A or Lodging Only:**

P(A or Lodging Only) = P(A) + P(Lodging Only) - P(A and Lodging Only)
P(A) = 110/300
P(Lodging Only) = 120/300
P(A and Lodging Only) = 30/300

P(A or Lodging Only) = (110/300) + (120/300) - (30/300) = 200/300 = 2/3

**(b) Probability of Hotel B with Breakfast, given Hotel B:**

This is a conditional probability. We want P(B with Breakfast | B).

P(B with Breakfast | B) = P(B with Breakfast and B) / P(B)
Since "B with Breakfast" is a subset of "B",  P(B with Breakfast and B) = P(B with Breakfast) = 120/300
P(B) = 210/300

P(B with Breakfast | B) = (120/300) / (210/300) = 120/210 = 4/7

**(c) Probability of Hotel A, given Lodging Only:**

This is also a conditional probability: P(A | Lodging Only)

P(A | Lodging Only) = P(A and Lodging Only) / P(Lodging Only)
P(A and Lodging Only) = 30/300
P(Lodging Only) = 120/300

P(A | Lodging Only) = (30/300) / (120/300) = 30/120 = 1/4

**(d) Probability of Hotel A without Breakfast or Hotel B with Breakfast:**

These are mutually exclusive events (they can't both happen at the same time), so we can simply add their probabilities.

P(A without Breakfast) = 30/300
P(B with Breakfast) = 120/300

P(A without Breakfast or B with Breakfast) = (30/300) + (120/300) = 150/300 = 1/2