You can count all the possibilities(sample space) for tossing a coin twice. Possibilities are: HH HT TT TH. So there are four possible outcomes and only one satisfies your demand i.e. HT. Hence probability of getting HT = 1/4.Now let’s solve it in more conceptual way. Suppose you toss a single coin then P(H) = 1/2 and P(T) = 1/2 assuming that coins is not biased. When you toss a coin twice, the second event is independent of first event. Therefore P(HT) = P(H).P(T) = 1/4.
