Computing Probability Suppose there are 10 students , out of which 6 are  selected. Assume I have a list of 5 students with me. What is the probability that AT LEAST 3 students from my list are among the 6 students selected?
I could not figure , this out ?? . Should I apply conditional probability ?
 A: You can just add up the chances that exactly three of your list are selected and three not on the list, the chance that exactly four of your list are selected...
A: Assuming that each person has the same probability of being chosen, this is just a counting problem involving combinations (not permutations). As Ross mentioned, we will be adding up the ways we can get exactly N of our 5 people selected, where N=3,4,5.
First, lets get the denominator, which is the total number of ways to select a group of 6 people from a group of 10: ${10 \choose 6}$.
Now, lets count the number of ways to get exactly 3 people from our list of 5 among the 6 chosen. First, we need to choose 3 people from our 5, which can be done $5 \choose 3$ ways. Now, we have to choose 3 people from the remaining 5 people not on our list, so again we have $5 \choose 3$ ways to do this. From the fundamental theorem of counting, we get a total of $ {5 \choose 3}^2$ ways to get exactly 3 of our people selected. 
In general, the probability that $N\geq 1$ people from your list of 5 are part of the group of 6 is: $P(N=n)= \frac{{5 \choose N}{5 \choose (6-N)}}{10 \choose 6}$ Also note that $P(N=0)=0$. Therefore, we can simplify our lives a bit by calculating $P(N\geq3)$ indirectly via $P(N\geq3)= 1 - [P(N=1)+P(N=2)]$ which gives us:
$P(N=1)=\frac{5\times1}{10 \choose 6}=\frac{5}{210}$
$P(N=2)=\frac{{5\choose2}{5\choose 4}}{10 \choose 6} = \frac{50}{210}$
Therefore, $P(N\geq3)=1-\frac{55}{210} \approx 74\%$
