Probability Problem with 10 players, Bob and a friend being on a team Please check my work.
Out of 10 players including Bob and his two best friends a team of 5 players will be formed. What is the likelihood of Bob making the team with at most one of his friends?
Total Possible Teams
$$\binom{10}{5} = 252$$
Bob with at least one other friend.
$$\binom{8}{3} = 56$$
Divide total possibilities with possibilities of teams with Bob and his friend.
$$\frac{\mbox{Combinations}}{\mbox{Total Possibilities}} = \frac{2}{9}$$
 A: There are indeed $$\binom{10}{5}$$ ways to form a team of five players from among the ten people.
If Bob is selected to be on the team, there are $$\binom{7}{4}$$ ways to select four teammates for Bob from among the seven people who are not his friends.
If Bob is selected to be on the team, there are $$\binom{2}{1}\binom{7}{3}$$ ways to select exactly one of his two friends and three of the other seven people to be on Bob's team.
Hence, the probability that Bob makes the team with at most one of his  friends is
$$\frac{\dbinom{7}{4} + \dbinom{2}{1}\dbinom{7}{3}}{\dbinom{10}{5}}$$
Note that the complementary event is that either Bob is not selected for the team, which can occur in $$\binom{9}{5}$$ ways since five of the other nine people must be selected to be on the team, or Bob and both his friends are selected, which can occur in $$\binom{2}{2}\binom{7}{2}$$ ways since both his friends and two of the other seven people must be selected to be on Bob's team.  Hence, the probability that Bob makes the team with at most one of his friends is $$1 - \frac{\dbinom{9}{5} + \dbinom{2}{2}\dbinom{7}{2}}{\dbinom{10}{5}}$$
Addendum:  In the comments, Graham Kemp suggested another method.  There are
$$\binom{9}{4}$$
teams which could include Bob since we must choose four of the other nine people to be his teammates.  We know that
$$\binom{2}{2}\binom{7}{2}$$
include both his friends.  Therefore, the number of teams which include at most one of his friends is
$$\binom{9}{4} - \binom{2}{2}\binom{7}{2}$$
Consequently, the probability that Bob makes the team with at most one of his friends is
$$\frac{\dbinom{9}{4} - \dbinom{2}{2}\dbinom{7}{2}}{\dbinom{10}{5}}$$
A: Your computation of total ways of forming the teams is correct, but that for favorable ways is not.
When learning a particular type of problem, it is best to avoid shortcuts and use a standard method.
The people involved are Bob (B), friend $F_1$, friend $F_2, 7\;$ others $\;(O)$
If exactly one friend is to be on the team along with Bob, [see PS] the selections which are "good" are $B,F_1,3O's\;$ and $B,F_2,3O's\;$
So using the formula for selection without replacement, you should be able to solve it.
Later on, when you gain more experience, you can think of how the process can be shortened.
PS $\;\;$Your header and body are contradictory. Header states "... Bob and a friend on the team..." whereas body states "... Bob and at most one of his friends.."
Taking the interpretation of the header to imply exactly one friend on the team along with Bob, the baby steps would be $\dfrac{\binom11\binom11\binom73 + \binom11\binom11\binom73}{\binom{12}5} = \dfrac5{18}$
It can, of course be simplified to $\dfrac{\binom21\binom73}{\binom{12}5}$
