Similarly, if one has a sample {"X"1, ..., "X""n"} then the probability that the next observation "X""n"+1 will be the largest is 1/("n" + 1), since all observations have equal probability of being the maximum. In the same way, the probability that "X""n"+1 will be the smallest is 1/("n" + 1). The other ("n" − 1)/("n" + 1) of the time, "X""n"+1 falls between the sample maximum and sample minimum of the sample {"X"1, ..., "X""n"}. Thus, denoting the sample maximum and minimum by "M" and "m," this yields an ("n" − 1)/("n" + 1) prediction interval of ["m", "M"].