SOLUTION: What is the probability of obtaining exacty seven heads in eight flips of a coin, given that at least one is a head.

P(7 heads|1 or more is a head) = P(7 heads AND 1 or more is a head)/P(1 or more is a head)

First we find the numerator:

P(7 heads AND 1 or more is a head)

The "AND 1 or more is a head" is redundant because if you have 7 heads you automatically have 1 or more heads,
so we just need the probability of 7 heads (out of 8 trials), which is

8C7(1/2)^7(1/2)1 = 8(1/2)^8 = 8(1/256) = 8/256 (we could reduce that but

we'll just leave it 8/256.

Next we find the denominator:
P(1 or more is a head)

We get the probability of the complement event which is to get no heads, which is

8C0(1/2)^0(1/2)^8 = 1(1/2)^8 = (1/2)^8 = 1/256

So the probability of the complement event is (1/2)^8, so the probability of the
event is 1-1/256 = 256/256-1/256 = 255/256


Now we go back to the original:

P(7 heads|1 or more is a head) = P(7 heads AND 1 or more is a head)/P(1 or more is a head)
 
(8/256)/(255/256) = (8/256)*(256/255) = 8/255

Edwin