You have 4 prizes, 3 tickets, n tickets- what is the probability of winning You have bought 3 tickets in a lottery. There are n total tickets and 4 prizes. What are the odds of winning at least one prize?
I thought of it like this:
The total possible ways of extracting 4 prizes is: a= $${n\choose 4}$$
The possibilities of extracting at least 1 winning prize is: b=$${4\choose 1} + {4\choose 2} +{4\choose 3}$$
so the probability of winning is $$\frac{{4\choose 1} + {4\choose 2} +{4\choose 3}}{{n\choose 4}}$$
If I'm wrong, which might be the case, please tell me what the right solution is, as this is the only I could come up with. Thank you
 A: We give several approaches. The first one is closest in spirit to yours. 
You have bought your $3$ tickets. Now the lottery corporation is choosing the $4$ winning tickets. These can be chosen in $\binom{n}{4}$ ways. If the lottery is well run, the process makes more or less sure that the choices are all equally likely. 
How many ways can they choose the $4$ winning tickets so that none of them are yours? Clearly $\binom{n-3}{4}$. So the probability you win no prize is 
$$\frac{\binom{n-3}{4}}{\binom{n}{4}}\tag{A}.$$
(The expresssion in (A) can be considerably simplified.) The probability you win at least one prize is therefore $1$ minus the answer of (A).
Another way: Imagine the lottery corporation picks the winning tickets  one at a time. The probability the first ticket it picks is bad (not one of yours) is $\frac{n-3}{n}$.
Given that the first ticket picked was bad, there are $n-4$ bad left out of $n-1$, So given that the first ticket was bad, the probability the second is bad is $\frac{n-4}{n-1}$. Thus the probability the first two are bad is 
$\frac{n-3}{n}\cdot\frac{n-4}{n-1}$.
Continue two more rounds. The same reasoning shows that the probability all our tickets are bad is 
$$\frac{n-3}{n}\cdot\frac{n-4}{n-1}\cdot\frac{n-5}{n-2}\cdot\frac{n-6}{n-3}.$$ 
Still another way: Let's switch points of view.
Imagine that the winning tickets have already been determined.  So there are $4$ "good" tickets and $n-4$ bad. We will find the probability that you pick all bad.
There are $\dbinom{n}{3}$ ways that we could choose our three tickets. There are $\binom{n-4}{3}$ ways to choose them so they are all bad. So the  probability that we choose all bad is 
$$\frac{\binom{n-4}{3}}{\binom{n}{3}}.\tag{B}$$
The probability of at least one good is $1$ minus the number in (B).
Remark: We could use a strategy like yours: Find the probability of winning exactly $1$ prize, exactly $2$ prizes, exactly $3$ prizes, and add up,
We calculate the probability of winning exactly one prize. The other calculations are roughly similar. So you have bought $3$ tickets. We find the probability exactly one is good,
There are $\binom{n}{4}$ ways for the corporation to choose  $4$ tickets. How many ways are there to choose $1$ that you have and $3$ that you don't have?
The one you have can be chosen in $\binom{3}{1}$ ways. For each such choice, the $3$ you don't have can be chosen in $\binom{n-3}{3}$ ways. Thus the probability of exactly one winning ticket is
$$\frac{\binom{3}{1}\binom{n-3}{3}}{\binom{n}{4}}.$$
A: You said that 
the possibilities of extracting at least 1 winning prize is: b=$${4\choose 1} + {4\choose 2} +{4\choose 3}$$
This is not true. 
When you're selecting $a$ winning tickets, where $0\leq a\leq 3$, you are also selecting $3-a$ losing tickets! And there are multiple ways of doing so! 
For example, the number of ways in which $1$ winning ticket can be selected is ${4\choose 1}{n-4\choose 2}$. 
Hence, the number of ways in which $1,2$ or $3$ tickets can be selected is$$= {4\choose 1}{n-4\choose 2}+{4\choose 2}{n-4\choose 1}+{4\choose 3}{n-4\choose 0}$$
The resultant probability is $$ \frac{{4\choose 1}{n-4\choose 2}+{4\choose 2}{n-4\choose 1}+{4\choose 1}{n-4\choose 0}}{{n\choose 3}}$$
