The proportion of defectives in a mass produced item is 5%. 
What is the probability that a random sample of 10 of these items will
contain exactly 2 defectives?

This is a binomial distribution problem.  In general, you're looking
for a set of 10 items where 2 of them are of the kind that has a 5%
probability.  That probability of such a set is, in general form:

(p^x)*[(1-p)^(n-x)] Where x=2, p=.05, n=10 for this problem.  

Now, since it doesn't matter what order these element come in, we need
to multiply that probability by the number of ways of rearranging the
set.  This is nCx where:

nCx= n!/(x!*(n-x)!)

Plugging in all the numbers gives:

10C2*(.02^.2)*(.95^8) = 

10!/(2!*8!)*(.05^2)*(.95^8) =

45*.00165855 = .0746 = 7.46% chance.

Hope this helps!

Thats great - thanks so much