Regarding probability I am a bit confused here.
I am playing a game, and at the end of each match, 3 random players out of a total of 10 are awarded prizes
What is the probability of me getting an item, considering any player may only get one item.
I'm sure it's not as easy as 3/10.
Thanks!
 A: Yes, it is that easy. (If the three winners are selected at random, any three equally likely.) 
But you could do it the hard way. There are $\binom{10}{3}$ ways to select $3$ winners. There are $\binom{9}{2}$ ways to select $2$ people to join you on the podium. So the probability is $\binom{9}{2}$ divided by $\binom{10}{3}$. After simplifying, you will get $3/10$.
To justify the easy way, suppose there are three medals awarded, gold, silver, and plastic. Let $A$ be the event you win the gold medal, $B$ be the event you win the silver, and $C$ the event you win the plastic. Then the events $A$, $B$, and $C$ are pairwise disjoint. I think you will see that the probability you win gold is $\frac{1}{10}$. We get the same values for silver and for plastic. For the event $A\cup B\cup C$ that you win some medal, add. 
A: Formally,
there are ${9 \choose 2}$ ways for you to get an item and ${10 \choose 3}$ ways for the three items to be distributed, so
$$
P(\text{you get an item})
= \frac{{9 \choose 2}}{{10 \choose 3}}
= \frac{9! 7! 3!}{10! 7! 2!} = \frac{3}{10}.
$$
However, there is an easier way to see this.
Simply notice that everyone has an equal chance of being picked to win a prize,
and the total expected number of prizes received by all payers is $3$.
Therefore each player must have contributed an expected value of $\frac{3}{10}$ to the total expected number of prizes.
