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Se escriben, cada uno en un papel, los dígitos desde el 1 al 9. Si se eligen al azar 2 papeles ¿Cual es la probabilidad de obtener como diferencia entre los dígitos el número 3?

User Zgpeace
by
7.2k points

1 Answer

7 votes
This can be solved by a contingency table to enumerate all possibilities.

1 2 3 4 5 6 7 8 9
1 - 1 2 3 4 5 6 7 8
2 1 - 1 2 3 4 5 6 7
3 2 1 - 1 2 3 4 5 6
4 3 2 1 - 1 2 3 4 5
5 4 3 2 1 - 1 2 3 4
6 5 4 3 2 1 - 1 2 3
7 6 5 4 3 2 1 - 1 2
8 7 6 5 4 3 2 1 - 1
9 8 7 6 5 4 3 2 1 -

The diagonal entries are void because we cannot pick two tags with the same number. We count 72 outcomes (excluding diagonal) and 12 of them are threes. Therefore
Probability that the difference is 3 = 12/72 = 1/6
User Jeff Klukas
by
6.5k points
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