What is the probability any one item in a set of 10 items is picked from a pool of 30? Consider that a set contains 30 distinct items. User must pick 10 distinct items.
What is the probability that any given item will appear in the set of items picked?
The probability that an item is picked from the set of 30 is 3.33%. I am assuming that because 10 items must be picked, the probability of any one item to appear in the final set is 33.33% (3.33% * 10 items), but I'm not sure if this is correct as the figure seems somewhat high. Is this right? 
 A: Yes, this is right. There are ${30 \choose 10}$ ways to pick $10$ distinct items from the set. If we pick any given item, there are ${29 \choose 9}$ ways to choose $9$ remaining items. So the probability that any given item will appear in the set of items picked is
$$
\frac{29 \choose 9}{30 \choose 10}=\frac13.
$$
Edit:
I think the approach $10\cdot\frac1{30}$ lacks a little more detailed explanation. The probability that the given item is picked first is $\frac1{30}$, the probability that the given item is picked second is $\frac{29}{30}\cdot\frac 1{29}$, the probability that a given item is picked third is $\frac{29}{30}\cdot\frac{28}{29}\cdot\frac{1}{28}$. So the probability that the given item will appear in the set of items picked is equal to
$$
\frac1{30}+\frac{29}{30}\cdot\frac 1{29}+\frac{29}{30}\cdot\frac{28}{29}\cdot\frac{1}{28}+\ldots=10\cdot\frac1{30}=\frac13.
$$
A: Your answer is correct. This has to do with the fact that 
$$Pr(\textrm{Picked)})=Pr(\bigcup_{i=1}^{30}\textrm{Picked in ith position})$$
which since each of these events are disjoint from one another we have 
$$Pr(\bigcup_{i=1}^{10}\textrm{Picked in ith position})=\sum_{i=1}^{10}Pr(\textrm{Picked in ith position})$$
where as you said that $Pr(\textrm{Picked in ith position})=\frac{1}{30}$ thus final answer is 
$$Pr(\textrm{Picked})=10\frac{1}{30}=\frac{1}{3}$$
A: There are 10 items picked; that's one third of all the items.   The special item could be any one of them, or it could be any one of the remaining 20.   So the probability of it being picked is one third.
Consider this.   The probability that it is the first item picked is: $1/30$.   The probability that it is the second item is also: $1/30$.   So the probability that it is either the first or second item is: $1/30 + 1/30$.   That is: $2/30$.   And so on, et cetera.
