Question 420908
<pre><font face = "consolas" size = 4 color = "indigo"><b>
The probability of selecting no bad ones is the probability of 
selecting 2 from the 8 good bulbs, which is C(8,2)/C(10,2) or as
some people write it (8C2)/(10C2) = 28/45

The probability of selecting 1 good one and one bad one, is the 
probability of selecting 1 from the 8 good bulbs, and one from 
the 2 bad ones, which is [C(8,1)C(2,1)]/C(10,2) or as some people 
write it (8C1*2C1)/(10C2) = 16/45.

The probability of selecting two bad one, is C(2,2)/C(10,2) or 
as some people write it (2C2)/(10C2) = 1/45.


The probability distribution function is:


x    p(x)
0   28/45
1   16/45
2    1/45

Notice that the sum of those three probabilities is

28/45 + 16/45 + 1/45 = 45/45 = 1.

The expectation is

E(x) = 0*(28/45) + 1*(16/45) + 2*(1/45) = 28/45 + 16/45 + 2/45 = 

46/45 = 1.022222

If we were to reach into the same box and draw out 2 bulbs 
thousands of times we would expect to average picking out 
1.02222 bad bulbs.  It doesn't mean that we would ever expect 
to draw out a fraction of a bulb!  That would be impossible.  
We just mean that we would expect to average 1.02222 bulbs 
if we were to repeat the experiment many times.

Edwin</pre>