From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?


C_n^k=\frac{n!}{k!(n-k)!}


The attempt at a solution

6 tickets from 10, we can choose on
C_1_0^6
.There are 4 tickets left.

C_7^3
is the tickets which are not winning. Is the right answer:



C_1_0^6
-
C_7^3

From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?
I'd start from the inverse.
How many combinations of six tickets are there altogether? You already have that right: C_{10}^6.

How many combinations of six tickets are there with no winning tickets? That would be 6 losing tickets out of 7.

Winning combinations = All combinations - Losing combinations

But why, 6 losing tickets out of 7?

You buy six tickets.
If all your tickets lose, then you have six of the seven losing tickets.

Yes. But six of seven, because three are for sure winning, righT?

Right. There are ten tickets altogether. Three are winners. Seven are losers.

So, the correct answer is:

C_10^6-C_7^6, right?

That's what I figure.

C_10^6-C_7^6=203

C_10^3-C_7^3=85

In this case, it says that it is better to buy 3 tickets, which doesn't make sense.

Six is better.
Three tickets gives you 85 winning combinations.
Six tickets gives you 203 winning combinations.

Yes. Sorry, and thank you for the help.

Shouldn't there be a 4 in there somewhere for the losing combinations ?

And the loser combinations should include 1, 2 or no winners because the winning set might only have one winner.

This answer is not clear to me yet.