Question

A typical ID number in a college class consists of a four-digit number, such as 0119. When an ID number is randomly selected, what is the probability that no two of its digits are the same

Answer 1

Answer:

$Probability = 0.504$

Step-by-step explanation:

Given

$ID: 4\ digit\ number$

Required

Determine the probability that no two digit is repeated

First, we need to determine the total possible number of pins (with repetition)

Each of the 4 ID digit can be selected from any of the 10 digits (0 - 9).

i.e.

$First = 10$

$Second = 10$

$Third = 10$

$Fourth = 10$

$Total = 10 * 10 * 10 * 10$

$Total = 10000$

i.e.

$n(Repetition) = 10000$

Next, we determine the total possible number of pins (without repetition)

Once a digit is selected from any of the 10 digits (0 - 9), it can no longer be repeated.

i.e.

$First = 10$

$Second = 10 - 1 = 9$

$Third = 10 - 1 - 1= 8$

$Fourth = 10 - 1 - 1 - 1 = 7$

$Total = 10 * 9 * 8 * 7$

$Total = 5040$

i.e.

$n(No\ Repetition) = 5040$

The probability is then calculated as:

$Probability = \frac{n(No\ Repetition)}{n(Repetition)}$

$Probability = \frac{5040}{10000}$

$Probability = 0.504$