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What is the probability that among the four digit numbers, to obtain a number whose two central digits are multiples of 3 and contains at least one digit of 5 value?

Thursday 2 January 2020,

What is the probability that among the four digit numbers, to obtain a number whose two central digits are multiples of 3 and contains at least one digit of 5 value?

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• How many four-digit numbers are there?

There are 10 possibilities for the digit (all digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), the same for the number of tens and hundreds. What happens to the unit of one thousand?

9x10x10x10 = 9000

There are 9,000 of 4 digit numbers.

The exercise considered four digit numbers where at least one of the numbers is 5, regardless of the two power stations (multiples of three). So we left two digits that may be five.

Consider cases in which there is no five and the central terms other than multiples of three: 8 x 7 x 7 x 9, the two core values represent the cases not multiples of 3 and the two extreme values not fives cases.

So 8 x there are in total 7 x 7 x 9 numbers that do not have at least one five and whose central terms are not multiples of 3.

Then the cases in which the central terms are multiples of three and the other digits contain at least one five are 9 x 10 x 10 x 10 - 8 x 7 x 7 x 9

Finally, the probability of obtaining the request is given by:

(9 x 10 x 10 x 10 - 8 x 7 x 7 x 9) /(9 x 10 x 10 x 10)

Cuántos números de cuatro dígitos existen?

Para el dígito de la unidad hay 10 posibilidades (todos los dígitos: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), lo mismo para la cifra de las decenas y centenas. ¿Qué ocurre con la cifra de la unidad de mil?

9x10x10x10 = 9000

Existen 9.000 números de 4 cifras.

El ejercicio considera números de cuatro cifras donde al menos una de las cifras sea 5, sin considerar los dos centrales (multiplos de tres). Así nos quedan dos dígitos que pueden ser cinco.

Formaremos la respuesta en forma indirecta:

Consideremos los casos en que no haya ningún cinco y que los términos centrales no sean multiplos de tres: 8 x 7 x 7 x 9, los dos valores centrales representan los casos no multiplos de 3 y los dos valores extremos los casos de no cincos.

Así en total hay 8 x 7 x7 x 9 números que no tienen al menos un cinco y cuyos terminos centrales no sean multiplos de 3.

Luego los casos en que los términos centrales sean multiplos de tres y los otros digitos contengan al menos un cinco son 9 x 10 x 10 x 10 - 8 x 7 x 7 x 9

(9 x 10 x10 x 10 - 8 x 7 x 7 x 9)/(9 x 10 x 10 x 10)

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