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.
2. Exactly five of your numbers are chosen.
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.
Now you know how to solve #3.