Question

A teacher randomly assigns her 30 students 4-digit I.D. numbers using the digits 0 to 9. No digits are repeated within an ID number. What is the probability one of those ID numbers is 9876?

Answer 1

`Probability = (1)/(5040)`

Explanation:

Given:

Number of students: 30

Student ID = 4 digit (0 - 9 without repetition within an ID)

Required:

Calculate the probability that one of those ID numbers is 9876?

Step 1: First we calculate the total possible IDs.

To do this, we make use of the permutation formula; We're using this formula because the question says assign

Hence, Number of possible IDs is as follows;

`^(n) P_(r)`

Where n = digit 0 - 9 = 10 digits

r = 4 digits

`^(n) P_(r)` becomes
`^(10) P_(4)`

`^(10) P_(4) = (10!)/((10 - 4)!)`

`^(10) P_(4) = (10!)/(6!)`

`^(10) P_(4) = (10 * 9 * 8 * 7 * 6!)/(6!)`

`^(10) P_(4) = 10 * 9 * 8 * 7`

`^(10) P_(4) = 5,040`

Step 2: Calculate the required probability

Probability is calculated as number of required outcomes divided by number of total outcomes

There are only 1 possibility of having a 9876 out of a total of 5,040

Hence, the probability of having one of those ID numbers as 9876 =

`Probability = (1)/(5040)`

Answer 2

The number of ways 4 digit numbers can be forms using the digits 0 to 9 with no number repeated is given by


`^(10)P_4= (10!)/((10-4)!) = (10!)/(6!) =10*9*8*7=5,040`

The probability that one of the numbers is 9876 is given by
`(1)/(5,040)`