Hello, mneox!

There are: . possible outcomes.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.

Right!

2.five of your numbers are chosen.Exactly

Among the 49 possible numbers, 6 are "winners", the other 43 are "losers".

We want 5 winners (from the available 6 winners)

. . and 1 loser (from the available 43 losers).

There are: . ways.

Therefore: .

Now you know how to solve #3.