Answer:
0.0143 = 1.43% probability that first 4 digits of the code are even numbers.
Step-by-step explanation:
Arrangements of n elements:
The number of possible arrangements of n elements is given by:
[tex]A_{n} = n![/tex]
Probability:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this question:
Desired outcomes:
First four digits are even(2, 4, 6 and 8 are even, so arrangements of 4 elements).
Last four digits are odd(1, 3, 5 and 7 are odd, so the last four elements are also arrangements of 4 elements). So
[tex]D = 4!*4! = 24*24 = 576[/tex]
Total outcomes:
Arrangements of 8 digits. So
[tex]T = A_{8} = 8! = 40320[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{576}{40320} = 0.0143[/tex]
0.0143 = 1.43% probability that first 4 digits of the code are even numbers.
