Probability of two Daniels in one group my professor gave us a "fun" problem to work on at home, but I am relatively new to probability. 
The question is as follows: There are four Daniels in a class of 42 students. If the class breaks into groups of three what is the probability that there is at least two Daniels in one group?
I am trying to solve the easier problem "What is the probability that there is two Daniels?" However, I am stuck even here. I will show that I have this far and why.
$$P(E)=\frac{{4 \choose 2}{40 \choose 1}}{{42\choose 3}}$$
I am using ${4 \choose 2}$ for choosing two daniels from 4, ${40 \choose 1}$, 1 student from the remaining forty, and  ${42\choose 3}$ for groups of 3 possible given 42 students. I may have a deep misunderstanding of probability, however I feel that the thing I feel most unsure about is using ${40 \choose 1}$. Part of me feels like it should be ${42 \choose 1}$, but I am not entirely sure why. 
Any help, or clarifications of things I may be confused about will be greatly appreciated.  
Edit after feedback: Attempted solution to original question$$P(E)=\frac{{4 \choose 2}{38 \choose 1}}{{42\choose 3}}+\frac{{4 \choose 3}}{{42\choose 3}}= \frac{29}{1435}$$

The $P(E)$ you've calculated is correct.
However, for solving the original problem, you'll have to calculate the probability of the event that there are $3$ Daniels in a group, and then add probabilities of both events.
Edit:
I re-read the problem, and your calculation seems a bit off.
For event $E$,
No of Daniels $= 4$;    ways of choosing $2$ from $4$ = $4 \choose 2$
No of "Non-Daniels" $= 42 - 4 = 38$; ways of choosing $1$ from $38$ = $38 \choose 1$.
Total groups of size $3 =$ ways of choosing $3$ from $42 =$ $42 \choose 3$
