The probability that all the numbers in the sequence are even is 12%
The count of number (n) in the lock is
[tex]n = 40[/tex] i.e. from 0 to 39
There are 20 even numbers from 0 to 39
Represent this with:
[tex]r = 20[/tex]
So, the probability that all three selected numbers (without repetition) are even numbers is:
[tex]Pr = \frac{r}{n} \times \frac{r - 1}{n - 1} \times \frac{r - 2}{n-2}[/tex]
This gives
[tex]Pr = \frac{20}{40} \times \frac{20 - 1}{40 - 1} \times \frac{20 - 2}{40-2}[/tex]
Evaluate all differences
[tex]Pr = \frac{20}{40} \times \frac{19}{39} \times \frac{18}{38}[/tex]
Reduce fractions
[tex]Pr = \frac{1}{2} \times \frac{1}{39} \times \frac{18}{2}[/tex]
Reduce fractions
[tex]Pr = \times \frac{1}{13} \times \frac{3}{2}[/tex]
So, we have:
[tex]Pr = \frac{3}{26}[/tex]
Divide
[tex]Pr = 0.115[/tex]
Express as percentage
[tex]Pr =11.5\%[/tex]
Approximate
[tex]Pr =12\%[/tex]
Hence, the probability that all the numbers in the sequence are even is 12%
Read more about probabilities at:
https://brainly.com/question/251701
