There are three passengers on an airport shuttle bus that makes stops at four different hotels. The probability that all three passengers will be staying at different hotels, is

Question

There are three passengers on an airport shuttle bus that makes stops at four different hotels. The probability that all three passengers will be staying at different hotels, is (A) (8/27) (B) (1/4) (C) (3/8) (D) (3/4)

Tardigrade Admin · Accepted Answer

The total number of differen choices is 4 · 4 · 4, while the number for which the three hotels are different 4 · 3 · 2

Q. There are three passengers on an airport shuttle bus that makes stops at four different hotels. The probability that all three passengers will be staying at different hotels, is

$\frac{8}{27}$

$\frac{1}{4}$

$\frac{3}{8}$

$\frac{3}{4}$

The total number of differen choices is 4 · 4 · 4, while the number for which the three hotels are different 4 · 3 · 2

Q. There are three passengers on an airport shuttle bus that makes stops at four different hotels. The probability that all three passengers will be staying at different hotels, is

278​

41​

83​

43​

The total number of differen choices is 4 · 4 · 4, while the number for which the three hotels are different 4 · 3 · 2

$\frac{8}{27}$

$\frac{1}{4}$

$\frac{3}{8}$

$\frac{3}{4}$