Question

four numbers from the set 0 to 9, including 0 and 9, are used to create a personal identification number. once a number is used, it cannot be reused. how many possible outcomes exist for the situation? how many possible outcomes begin with a 0? what is the theoretical probability of choosing a personal identification number that begins with a 0?

Explanation:

Step1: Calculate total number of outcomes

The number of permutations of $n$ distinct objects taken $r$ at a time is given by $P(n,r)=\frac{n!}{(n - r)!}$. Here, $n = 10$ (numbers from 0 - 9) and $r=4$. So, $P(10,4)=\frac{10!}{(10 - 4)!}=\frac{10!}{6!}=10\times9\times8\times7 = 5040$.

Step2: Calculate number of outcomes starting with 0

If the first - digit is 0, then we need to choose 3 numbers from the remaining 9 numbers. Using the permutation formula with $n = 9$ and $r = 3$, we have $P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7=504$.

Step3: Calculate the probability

The probability $P$ of an event is given by $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So, $P=\frac{504}{5040}=\frac{1}{10}=0.1$.

Answer:

How many possible outcomes exist for the situation? 5040
How many possible outcomes begin with a 0? 504
What is the theoretical probability of choosing a personal identification number that begins with a 0? 0.1