Question

(5). (10 points) There are 5 hotels in Stony Brook. If 3 people check into hotels on September 12, what is the probability that they each check into a different hotel? (What assumptions are you making?) Make sure to define any notation you use to describe elements of the sample space.

Answer 1

Answer:

0.48

Step-by-step explanation:

Probability that the first person chooses a hotel

⁵C₁

$^5C_1=\frac{5!}{(5-1)!1!}\\=\frac{120}{24}=5$

Probability that the second person chooses a different hotel

⁴C₁

$^4C_1=\frac{4!}{(4-1)!1!}\\=\frac{24}{6}=4$

because the choice of hotels has reduced by 1 as one hotel is occupied by the first person

Probability that the second person chooses a different hotel

³C₁

$^3C_1=\frac{3!}{(3-1)!1!}\\=\frac{6}{2}=3$

because the choice of hotels has reduced by 2 as two different hotels are occupied by the first person and second person

∴ The favorable outcomes are =⁵C₁×⁴C₁׳C₁=5×4×3=60

The total number of outcomes=5³=125

∴Probability that they each check into a different hotel=60/125=0.48