1) Let's find the sample space, i.e. the number of the total outcomes (favorable and not favorable). Let's consider that we need to pick 2 out of ten, that is our sample space.
[tex]_{10}C_2=\frac{10!}{2!\times8!}=45[/tex]2) Now, let's find the number of favorable events. In this case, both players must be a forward. The favorable event includes only forward players out of those 6:
[tex]_6C_2=\frac{6!}{2!(6-2)!}=15[/tex]3) We can write out the probability of picking one forward and another forward. Note that there was no replacement after each picking:
[tex]P=\frac{15}{45}=\frac{1}{3}[/tex]