Using the combination formula, it is found that there is a 0.7778 = 77.78% probability of choosing at least one even number.
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, there are 10 numbers numbered from 1 to 10.
The total number of outcomes is given by:
[tex]T = C_{10,2} = \frac{10!}{2!8!} = 45[/tex]
The number of outcomes with only odd numbers is given by:
[tex]O = C_{5,2} = \frac{5!}{2!3!} = 10[/tex]
Hence 45 - 10 = 35 outcomes have at least one even number, so the probability is:
p = 35/45 = 0.7778.
More can be learned about the combination formula at https://brainly.com/question/25821700
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