$14 \cdot 23$ A die is rolled six times. The probability of getting at least 5 on at least five of the rolls is
(A) $\frac{13}{729}$.
(B) $\frac{12}{729}$.
(C) $\frac{2}{729}$.
(D) $\frac{3}{729}$.
(E) None of these.
(25th American High School Mathematics Examination, 1974)
[Solution] Let $A$ represent the event of rolling at least a five, then the probability of $A$ is
$$
\frac{2}{6}=\frac{1}{3} \text {. }
$$

In six rolls of the die, the probability of event $A$ occurring six times is $\left(\frac{1}{3}\right)^{6}$.
In a specific sequence, the probability of event $A$ occurring five times and not occurring once is
$$
\left(\frac{1}{3}\right)^{5} \cdot \frac{2}{3} \text {. }
$$

Since there are six specific sequences for this to happen, the probability of $A$ occurring five times and not occurring once is
$$
6\left(\frac{1}{3}\right)^{5} \cdot \frac{2}{3}=\frac{12}{729} \text {. }
$$

Therefore, the probability of $A$ occurring all six times or five times with one non-occurrence is
$$
\frac{12}{729}+\left(\frac{1}{3}\right)^{6}=\frac{13}{729} \text {. }
$$

Hence, the answer is (A).