1. ## Probability with Combinatorics

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

2. Hello, mneox!

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!
There are: .$\displaystyle _{49}C_6$ possible outcomes.

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: .$\displaystyle (_6C_5)(_{43}C_1)$ ways.

Therefore: .$\displaystyle P(\text{exactly 5 winners}) \;=\;\frac{(_6C_5)(_{43}C_1)}{_{49}C_6}$

Now you know how to solve #3.