SOLUTION: Suppose a fair coin is tossed 12 times. How many different sequences of heads and tails can result? What is the probability that it will land heads exactly one time? What

Choose the 1st toss either of 2 ways, {heads,tails}
Choose the 2nd toss either of 2 ways, {heads,tails}
Choose the 3rd toss either of 2 ways, {heads,tails}
Choose the 4th toss either of 2 ways, {heads,tails}
Choose the 5th toss either of 2 ways, {heads,tails}
Choose the 6th toss either of 2 ways, {heads,tails}
Choose the 7th toss either of 2 ways, {heads,tails}
Choose the 8th toss either of 2 ways, {heads,tails}
Choose the 9th toss either of 2 ways, {heads,tails}
Choose the 10th toss either of 2 ways, {heads,tails}
Choose the 11th toss either of 2 ways, {heads,tails}
Choose the 12th toss either of 2 ways, {heads,tails}

That's 2×2×2×2×2×2×2×2×2×2×2×2 = 212 = 4096

Choose which toss will be the 1 head (the rest tails) in 12 ways.
So the probability is 12 ways out of 4096, or 12/4096, which reduces to 3/1024.

The complement event is that it will land tails all 12 times.

The probability of "all tails" is 1 way out of 4096 or 1/4096

So the probability of at least one head is 1 - 1/4096 or 4095/4096.