63% of people are happy. Pick 10 people randomly.
a) What are the odds that 8 of them are happy?
b) What are the odds that at least 2 of them are happy?
Where did you get this problem? It is strange that you should be given such a problem without having learned anything about problems like this. As romsek said, this is a pretty standard "binomial distribution" problem. Do you know what that is? Even if you do not, it is not too difficult to 'reason it out'. The probability that the first 8 people chosen are happy is $\displaystyle 0.63^8$. The probability the other two are not is $\displaystyle 0.37^2$. The probability the first 8 are happy and the last two are not is $\displaystyle (0.63^8)(0.37^2)$. In fact, it is not difficult to see that the probability that 8 are happy and 2 are not in any specific order is also $\displaystyle (0.63^8)(0.37^2)$. But there are 10!/8!2!= 45 different orders.
To find the probability that "at least 2" are happy, find the probability that none are happy and that 1 is happy, add to find the probability that either none or 1 is happy then subtract from 1.