When flipping a coin, is the probability of at least two tails complementary to at most one tail, if flipped three times in total? I got an answer on a Stats test wrong and I can't wrap my head around this. 
The solutions provided by the teacher: http://imgur.com/a/Ku1bt states that there is 50% probability of getting at least two tails out of three flips of a coin while the probability of getting at most one tail is 25%. 
I can't visualize how that could be possible. My calculations also belie this...
 A: You're right of course, the two probabilities must add to one. Both probabilities are actually 50%. As @mark999 has already noted in a comment, the solution (c) provided by your teacher has omitted to count the outcomes HTH and HHT.
One can find the probabilities of these events very quickly without enumerating all the possibilities. Three is an odd number of tosses, so the number of heads and tails in three tosses cannot be equal. One of them must occur more often than the other. By symmetry, heads and tails have an equal chance of being the more numerous. So we must have Pr(at least 2 heads) = Pr(at least two tails) = 0.5.
In general, if you flip a fair coin $n$ times, where $n$ is odd, then the probability that more than half the outcomes are heads [$(n+1)/2$ or more heads] is 0.5. And the probability that less than half the outcomes are tails [$(n-1)/2$ or fewer tails] is also 0.5, in fact this is the same event.
A: Your teacher's answer to the second part is wrong. The probability of getting at most one tail is also 50%. The complementary of at least two tails is at most one head in this case(not at most one tail, if you have at least two tails, how can you have at most one tail?). However, the probability of at most one head is the same as the probability of at most one tail. Does this make sense?
