umaa wrote:[spoiler]For the first question, it says we should multiply all the chances. ie, 2*2*2*2*2*2. Why didn't we take factorial? I'm pretty weak @ Prob, perm & combination. So, any valid explanation is welcome.[/spoiler]

If we wanted to, we could use factorials (in the form of combinations).
a) The total number of configurations could counted as follows:
(The total number of configurations with zero lights on) + (The total number of configurations with one light on) + (The total number of configurations with two lights on) + . . . and so on.
Since the order in which the lights are on doesn't matter (e.g., having lights 1, 2 and 5 on is the same as have lights 5, 1 and 2 on), we can use combinations.
So, we get: 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 which equals 1+6+15+20+15+6+1=64
Another way to calculate this is to break the counting task into stages and apply the Fundamental Counting Principle.
We get (the number of ways in which light 1 can be configured) x (the number of ways in which light 2 can be configured) x (the number of ways in which light 3 can be configured) x etc.
Since each light can be either on or off, each stage can be accomplished in 2 stages, so we get 2x2x2x2x2x2 = 64

b) The number of way in which 3 bulbs can be lit. We have 6 bulbs and we want to select 3 to be lit. Since order doesn't matter, this can be accomplished in 6C3 ways (20 ways)

c) probability = (# of outcomes favorable to having 3 lit bulbs)/(total # of outcomes) = 20/36 = 5/9

umaa wrote:Thanks Brent.
By the way, for the third question isn't it 20/64?

Oops, yes it's 20/64 (5/16)
I had a hard time simply combining answers from parts a) and b)