Problem 408

Suppose that when the three dice are rolled, the numbers rolled
are $x$, $y$ and $z$.

Since there are 6 possibilities for each of $x$, $y$
and $z$, there are $6 \cdot 6 \cdot 6 = 216$ possible
outcomes.

Also, the sum, $S$, of the three
rolls is at least $3 \cdot 1 = 3$ and
at most $3 \cdot 6 = 18$.

The outcome "$S > 5$" is the
complement of the outcome "$S $\leq$
5$".

Thus, the probability that $S>5$
is $1$ minus the probability that
$S \leq 5$.

It is easier to compute the probability that $S \leq 5$ directly by listing the rolls
that give this.

If $S = 3$, then $x+y+z=3$ and so $(x,y,z)=(1,1,1)$.

If $S = 4$, then $x+y+z=4$ and so $x$, $y$
and $z$ must be $1$, $1$
and $2$ in some order. Thus, $(x,y,z) = (2,1,1)$ or $(1,2,1)$ or $(1,1,2)$.

If $S = 5$, then $x+y+z=5$ and so $x$, $y$
and $z$ must be $1$, $1$
and $3$, or $1$, $2$
and $2$ in some order. There are
$3$ arrangements in each case and so
$6$ triples in total.

Therefore, there are $1 + 3 + 6 = 10$
triples with $S \leq 5$, and so the
probability that $S > 5$ is equal
to $1 - $\frac{10}{216} \approx$
0.954$.

Of the given choices, this is closest to $0.95$.