“Si se elige uno de los que…” cambia el denominador: solo cuenta ese grupo.
Estrategia paso a paso
Rodea en la tabla la fila o columna del grupo condicionante antes de dividir.
Trampas habituales
Usar el total general como denominador en una probabilidad condicionada.
Ejemplo resuelto
De 30 alumnos que estudian francés, 12 son de 10.º → P(10.º, sabiendo que estudia francés) = $\frac{12}{30}=\frac{2}{5}$.
Preguntas de ejemplo
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
Pregunta 1
Dificultad fácil≈ 1 min
A bag contains $4$ red marbles, $6$ blue marbles, and $5$ green marbles. If one marble is selected from the bag at random, what is the probability that it is blue?
AA. $\frac{1}{15}$
BB. $\frac{1}{3}$
CC. $\frac{2}{5}$
DD. $\frac{2}{3}$
Ver respuesta y explicación
Respuesta correcta: C. $\frac{2}{5}$
Explicación
There are $4+6+5=15$ marbles, and $6$ are blue, so the probability is $\frac{6}{15}=\frac{2}{5}$.
$\frac{1}{3}$ treats the three colors as equally likely. $\frac{2}{3}$ compares blue marbles with the $9$ marbles that are not blue. $\frac{1}{15}$ is the probability of selecting one particular marble.
Pregunta 2
Dificultad media≈ 1 min
An integer from $1$ to $30$, inclusive, is selected at random. What is the probability that the integer is a multiple of $4$ or a multiple of $6$, or both?
Respuesta numérica (escribes el número).
Ver respuesta y explicación
Respuesta: 1/3 · .3333
Explicación
Multiples of $4$: $4, 8, 12, 16, 20, 24, 28$ ($7$ numbers). Multiples of $6$: $6, 12, 18, 24, 30$ ($5$ numbers). The numbers $12$ and $24$ are in both lists, so there are $7+5-2=10$ numbers. The probability is $\frac{10}{30}=\frac{1}{3}$.
Counting $12$ and $24$ twice gives $\frac{12}{30}$, which is too large.
Pregunta 3
Dificultad difícil≈ 2 min
A box contains $5$ red balls and $3$ white balls. Two balls are selected from the box at random, one after the other, without replacement. What is the probability that both balls are red?
AA. $\frac{5}{14}$
BB. $\frac{25}{64}$
CC. $\frac{15}{28}$
DD. $\frac{5}{8}$
Ver respuesta y explicación
Respuesta correcta: A. $\frac{5}{14}$
Explicación
The probability that the first ball is red is $\frac{5}{8}$. After one red ball is removed, $4$ of the remaining $7$ balls are red, so the probability that the second is red is $\frac{4}{7}$. Both: $\frac{5}{8}\cdot\frac{4}{7}=\frac{20}{56}=\frac{5}{14}$.
$\frac{25}{64}$ assumes the first ball is put back. $\frac{15}{28}$ is the probability of one red ball and one white ball. $\frac{5}{8}$ is the probability for the first ball only.