A standard die with six faces is tossed onto a table. Itai counts the total number of dots on the five faces that are not lying on the table. What is the probability that this total is at least 19 ?
The total number of dots on the six faces of a standard six-faced die is $1+2+3+4+5+6=21$. When one face is lying on a table, the total number of dots visible equals 21 minus the number of dots on the face that is lying on the table.

For this total to be at least 19, then the number of dots on the face lying on the table must be 1 or 2 .

Therefore, the total is at least 19 when either of 2 of the 6 faces is lying on the table.

The probability of this is $\frac{2}{6}$ which equals $\frac{1}{3}$.

ANSWER: $\frac{1}{3}$