Probability Of People Visiting Pubs 
5 people went out to pubs. everyone individually choose a pub randomly from 10 pubs available.
  What is the probability that at "Pub1" "Pub2" came at least one person?

$|\Omega|=10^5$  
How can I look at this question from set point of view? (e.g $(A_1 \cap A_2)\cup(A_3\cup A_4 \cup A_5)$)
What is the problem with the following answer $\frac{5*4*10^3}{10^5}$?
 A: To frame things your way, let $A_i$ be the event of having someone visit bar number $i$. We want to count $$|A_1|+|A_2|- |A_1 \cap A_2| $$ , i.e., this is the number of ways we can have someone visit bar1 or bar2 , but not both. To get the total, we substract this from the total number of events
We have , for $A_1$:
There are $5$ choices for a person visiting bar1. Then there are $10^4$ ways for the remaining $4$ people to visit the $10$ bars, for a total of $5.10^4=50000$ ways. 
For $A_2$: Same thing: we choose a person in $5$ ways to visit bar $2$, and then the remaining $4$ people can visit the remaining bars in $10^4$ ways, for a total of $5.10^4 =50000$ ways.
For $A_1 \cap A_2. $ This is the number of ways of having at least one person in each bar. We can choose in $5C2$ ways the $2$ people so that there is one person in each of bar1, bar2. Then there are $3$ people left, who can choose the remaining $10$ bars in $10^3$ ways, for a total of $5C2. 10^3= 10^4=10000$ ways.  
So the total is $50000+50000-10000=90000$ ways.
So the probability is $\frac {10^5-90000}{10^5}=0.1=10$%
A: What is the probability it didn't happen?
It is equal to the probability at least one of the two bars where empty. Let $A$ be the set of outcomes in which pub $1$ is empty and $B$ the set of outcomes in which pub $2$ is empty. We want:
$P(A)+P(B)-P(A\cap B)$.
What is $P(A)$?
There are $9^5$ ways in which bar $1$ is empty.
What is $P(B)$? there are $9^5$ ways in which bar $2$ is empty.
What is $P(A\cap B)$?
There are $8^5$ ways in which both are empty.
Hence the probability is $\frac{9^5+9^5-8^5}{10^5}=\frac{85330}{10^5}$.
The probability we are looking for is therefore $\frac{100000-85330}{10^5}=\frac{14670}{100000}=0.14670$
A: P( choosing at least one to pub1,pub2) =1- p( no one choosing pub1, pub2)


*

*both pubs are not chooses so remaining 8 pubs are chooses by 5 persons randomly in 8^5 ways 

*total pubs 10 chosen by 5 people in 10^5 ways 


So probability 
P=1- [8^5/10^5]
P=1-[32,768/ 100,000]
P=0.67232
Answer
