Probability of selecting object from an urn Let's say I have an urn with 10 unique objects, and I choose 3 objects from it (each choice is made without replacement). Then the probability of choosing any one object is 3/10. I calculated this probability by summing the probability the object is chosen on the first pick + probability chosen on second pick + probability it's chosen on the third pick. I'm wondering if there is a simpler/more intuitive way to arrive at this probability though.
One idea I had is that for any random permutation of the 10 objects, we can split the permutation into first a group of 3 and second a group of 7 objects. There's then a 3/10 probability that a given object is in the first group (first set of 3 objects chosen).
 A: For the first trial, probability equals 1/10
For success in the second trial, probability = (1/9) * (9/10) = 1/10,  the first fraction gives the probability of choosing  the given object from the remaining 9 objects( since the tenth object has been removed from the first trial)
For success in the third trial, (1/8) * (8/9) * (9/10) = 1/10
Here the first fraction comes from the probability of choosing the given object from the remaining 8. The third fraction  comes from the fact that initially there are 10 objects and 9 ways to make the wrong choice. The second fraction is because after the first trial, there are 8 ways out of 9 to make the wrong selection again.
Hence, total probability = 3/10.
A: First, see Pascal's Triangle.
The stated problem can be attacked combinatorically.  Although I don't recommend it for this specific problem, for more complicated Probability problems, the combinatoric approach is a nice weapon to have.
The probability will be expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where $D$ is the total number of ways of selecting any $(3)$ items, and $N$ is the total number of ways of selecting $(3)$ items, one of which is the desired item.
Then:

*

*$D = \binom{10}{3} = 120$. 
Here, there are $120$ total ways of selecting any $(3)$ items from $(10)$ items, selected without replacement, where the order that the items are selected is regarded as not relevant.


*$N = \binom{9}{2} = 36.$ 
Here, assuming that one of the items selected is the desired item, there are $(36)$ ways of selecting the other $(2)$ items, from the remaining $(9)$ items.
Therefore, the probability that the desired item will be selected is
$$\frac{\binom{9}{2}}{\binom{10}{3}} = \frac{36}{120} = \frac{3}{10}.$$
A: The order is not important here, you simply choose $3$ elements from the $10$ offered:

The number of all possibilities is therefore $\binom {10} 3$.
To count the number of all desired triplets, i.e. those containing the desired element, we realize that to obtain such a triplet, we have

*

*$\binom 1 1$ possibilities for selecting $1$ desired element from the $1$ offered:



*$\binom 9 2$ possibilities for selecting $2$ other elements from the $9$ offered:
  
That is, there are $\binom 1 1 \times \binom 9 2$ possibilities to choose a desired triplet, consisting of $1$ desired and $2$ other elements.

Conclusion:
$$P(\text {chosing the desired element}) = {{\binom 1 1 \cdot \binom 9 2} \over {\binom {10} 3}} = {3 \over {10}}$$
