Probability that the lock code consists of all even digits is [tex]1344[/tex] if we choose the [tex]4[/tex] digit lock is randomly from numbers [tex]1,2,3,4,5,6,7,8,9[/tex]
How can we find the probability that the lock code consists of all even digits?
Total numbers are [tex]1,2,3,4,5,6,7,8,9[/tex]
And we choose [tex]4[/tex] digit code
So probability of choosing [tex]4[/tex] numbers from [tex]9[/tex] numbers without reapitation is
[tex]9p_{4} =\frac{9!}{(9-4)!} \\=\frac{9*8*7*6*5!}{5!} \\=3024[/tex]
For All numbers are even , probability the last digit of number is even.
So we have [tex]4[/tex] even number and if we fix one on last digit and can change other [tex]3[/tex] then the probability is
[tex]4*8P_{3} =4*\frac{8!}{(8-3)!} \\=4*\frac{8*7*6*5!}{5!} \\=1344[/tex]
So the total probability of code lock having all even number is [tex]1344[/tex]
Learn more about the probability is here :
https://brainly.com/question/28048708
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