Question

You want to make five-lener codes that use the letters AFER, and M without repeating any lener What is the probability that a randomly chosen code

starts with M and ends with E?
O A
001
OB. 002
OC 005
OD. 010

Answer 1

There are 5 letters to choose from for the first letter, but since the code must start with M, there is only 1 choice for the first letter. There are then 4 letters to choose from for the second letter, 3 for the third letter, 2 for the fourth letter, and 1 for the fifth letter. Therefore, there are:

1 x 4 x 3 x 2 x 1 = 24

possible codes that can be made.

Out of these 24 codes, there is only 1 code that starts with M and ends with E: MAFEA. Therefore, the probability of randomly choosing a code that starts with M and ends with E is:

1/24

Answer: A. 001.

Answer 2

`|\Omega|=5!=120\\|A|=3!=6\\\\P(A)=(6)/(120)=(1)/(20)=0.05`