## Task 35/78

In the USSR, it is customary to check in urban public transport, after tearing off the ticket control section, whether one has a "lucky ticket." This is the case when the sum of the first three digits equals the sum of the last three digits of the six-digit control number.

What is the probability that a "lucky ticket" has a sum of 7?
The number 7 can be decomposed into three summands in exactly eight different ways:

$$
\begin{array}{llll}
7+0+0(1) & 6+1+0(2) & 5+2+0(3) & 5+1+1(4) \\
4+3+0(5) & 6+2+1(6) & 3+3+1(7) & 3+2+2(8)
\end{array}
$$

The (unordered) triples (1), (4), (7), and (8) each contain a repeated digit, so for each of them, it is possible to form three ordered triples. For the (unordered) triples (2), (3), (5), and (6), six ordered triples are possible. Consequently, there are a total of \(3 \cdot 4 + 6 \cdot 4 = 36\) ordered triples with a sum of 7.

Since there are 1000 possible combinations of three-digit numbers (000...999), the probability of a sum of 7 in one of these combinations is 0.036.

Since the event of a "lucky ticket" with a sum of 7 is a compound event consisting of the logical product of two independent (identical) sub-events, its probability is equal to the square of the probability of each sub-event, i.e., \(0.036^2 = 0.001296\).