(c) (6 marks) I randomly chose a 3-digit code for my bike lock, only considering codes with 3 different digits.
El How many different codes exist? [I How many different codes exist with last digit 9? El What is the
probability that the last digit of my randomly chosen code is a 9?

solution: 

a) N=720___Total number of different 3-digit codes that can be formed without repetition.

b) n(A)=72___different codes that end in the digit 9.

c) P(A)=0.10___probability that the last digit of my randomly chosen code is a 9.

Step-by-step explanation

we have a total of nine digits: 0-1-2-3-4-5-6-7-8-9

we want to form a code of 3 different digits.
This is an activity in 3 steps, for the first step there are 10 digits available, for the second step 9 and for the third step there are 8 digits left. Then the multiplication rule applies.

a) N=10∗9∗8=720__Total number of different 3-digit codes that can be formed without repetition.

b) A: event that the last digit is 9.

considering that the last digit must be 9, and that repetition is not valid. This is a 2-step activity: step 1 select the first digit of 9 possible, then step 2 select the second digit of the remaining 8. then multiplication rule.

n(A)=9∗8=72____different codes that end in the digit 9.

c) P(A)=Nn(A)?=72072?=101?=0.10___probability that the last digit of my randomly chosen code is a 9.

bibliography: probability and statistics by george canavos.