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?

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

[tex] ^{10}P_4= \frac{10!}{(10-4)!} = \frac{10!}{6!} =10\times9\times8\times7=5,040[/tex]

The probability that one of the numbers is 9876 is given by [tex] \frac{1}{5,040} [/tex]

Answer Link

MrRoyal MrRoyal · Answer 2 · 2020-04-26T02:50:53+02:00

Answer:

[tex]Probability = \frac{1}{5040}[/tex]

Step-by-step 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;

[tex]^{n} P_{r}[/tex]

Where n = digit 0 - 9 = 10 digits

r = 4 digits

[tex]^{n} P_{r}[/tex] becomes [tex]^{10} P_{4}[/tex]

[tex]^{10} P_{4} = \frac{10!}{(10 - 4)!}[/tex]

[tex]^{10} P_{4} = \frac{10!}{6!}[/tex]

[tex]^{10} P_{4} = \frac{10 * 9 * 8 * 7 * 6!}{6!}[/tex]

[tex]^{10} P_{4} = 10 * 9 * 8 * 7[/tex]

[tex]^{10} P_{4} = 5,040[/tex]

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 =

[tex]Probability = \frac{1}{5040}[/tex]