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qaws [65]
2 years ago
15

En una fiesta hay 40 asistentes de los cuales el número de varones que no bailan es el doble que el número de las mujeres que no

bailan en un determinado momento. Si en ese instante se escoge un varón al azar, ¿cuál es la probabilidad que no este bailando si 5 mujeres están bailando?
Mathematics
1 answer:
Arte-miy333 [17]2 years ago
6 0

Answer:

50%

Step-by-step explanation:

Para resolver este problema vamos a definir 4 variables:

V= número de varones bailando.

M= número de mujeres bailando.

V' = número de varones que no están bailando.

M' = número de mujeres que no están bailando.

Sabemos que hay 40 personas en la fiesta, por lo que podemos construir la siguiente ecuación:

V+M+V'+M'=40

Vamos a suponer que cada mujer que está bailando, está bailando varón respectivamente. (El problema no da más información, entonces podemos suponer esto.)

Entonces si hay 5 mujeres bailando, entonces también hay 5 hombres bailando, por lo que nuestra ecuación se reescribe de la siguiente manera:

5+5+V'+M'=40

y simplificamos:

10+V'+M'=40

V'+M'=40-10

V'+M'=30

Ahora bien, el problem nos dice que el número de varones que no bailan es el doble del número de mujeres que no bailan en un determinado momento. Entonces con esta información podemos construir la siguiente ecuación:

V'=2M'

Y podemos despejar el número de mujeres que no bailan, lo que nos da:

M'=\frac{V'}{2}

Entonces podemos sustituir esto dentro de nuestra ecuación para obtener:

\frac{V'}{2}+V'=30

y podemos entonces despejar V'

\frac{3V'}{2}=30

V'=\frac{2(30)}{3}

V'=20

Entonces hay 20 varones que no están bailando, por lo que la probabilidad de que el varón que se escoge al azar no esté bailando está dada por la siguiente fórmula:

P=\frac{V'}{total}

P=\frac{20}{40}=\frac{1}{2}

P=50%

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