Probability of 2 or more students from a group of 5 being on the same team of 6 in a class of 36

To clarify the title: There is a class of 36 which is to be divided into 6 teams of 6. There are 5 specific students of interest, and I want to figure out the probability of 2 or more of those 5 being on the same team.

I know there are other questions like this, but I’ve tried to use multiple methods to do this and keep getting different answers so I’m clearly doing something wrong.

I started by trying this:

$$\frac{6*5*4*3*2}{36*35*34*33*32} + \frac{6*5*4*3}{36*35*34*33} + \frac{6*5*4}{36*35*34} + \frac{6*5}{36*35}$$

I got around 19.35%.

Then, I tried this because I thought maybe the first team member can go anywhere so the 6/36 is unnecessary:

$$\frac{5*4*3*2}{35*34*33*32} + \frac{5*4*3}{35*34*33} + \frac{5*4}{35*34} + \frac{5}{35}$$

I got 16.13%.

Then, I tried to modify a method I saw in a similar answer like so:

$$\sum_{n=2}^{5} \frac{_{36-n}C_{6-n}}{_{36}C_{6}}$$

I got 2.69%.

Yeah, I really don't remember how this works...

It appears that you want the probability that the specified students are not all on different teams. This is simply the compliment $(1-p)$ of the probability that they are all on different teams.

$$P(\text{all different teams})=\frac{36}{36}\times\frac{30}{35}\times\frac{24}{34}\times\frac{18}{33}\times\frac{12}{32}=\frac{162}{1309}\approx12.4\%$$

$$P(\text{not all different teams})=1-\frac{162}{1309}=\frac{1147}{1309}\approx87.6\%$$