Problem B. 3835. (September 2005)

B. 3835. This spring there were three Hungarian teams among the best eight in the EHF women's handball championship. When the eight teams were paired randomly, each Hungarian team found that they were playing an opponent from a foreign country. What was the probability of this?

(3 pont)

Deadline expired on October 17, 2005.

Solution 1. It is easy to check that the number of all possible matchings of the 8 teams is 7^.5^.3^.1=105.

The number of those cases when the opponents of the Hungarian teams are all foreign is 5^.4^.3=60.

The requested probability is $\frac{60}{105}=\frac47$ .

Solution 2. Denote the three Hungarian teams by A, B and C. Team A has 7 possible opponents. Therefore, the probablility that the opponent of team A is team B, is exactly 1/7.

The same holds for the pairs A,C and B,C.

The three wrong cases are pairwise distinct and all remaining cases are good. Therefore the requested probablity is

$p=1-3\cdot\frac17=\frac47.$

Statistics:

394 students sent a solution.

3 points: 217 students.

2 points: 29 students.

1 point: 66 students.

0 point: 79 students.

Unfair, not evaluated: 3 solutionss.

Problems in Mathematics of KöMaL, September 2005

394 students sent a solution.
3 points:	217 students.
2 points:	29 students.
1 point:	66 students.
0 point:	79 students.
Unfair, not evaluated:	3 solutionss.

Already signed up?
New to KöMaL?