if the lottery had 49 numbers you could choose, and you have to pick 6, what are the chances that:

1. All your numbers are chosen

2. 5 of your numbers are chosen

3. 4 of your numbers are chosen

for #1, i did 1/(49C6), is that the correct way to do it? my reasoning behind it is that there's 49C6 combinations one could choose, and only 1 set of those are what you have, and what could get picked.

i'm really confused for questions 2 and 3, so any help would be appreciated thx