ACT · Math · Preparing for Higher Math
Media, mediana, probabilidad simple y condicionada, y valor esperado.
Media, mediana, tablas de doble entrada, probabilidad y valor esperado.
P(2 rojas de $5R$ y $3A$, sin reposición) = $\frac{5}{8}\cdot\frac{4}{7}=\frac{5}{14}$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
What is the mean of the data set below?
4, 7, 7, 9, 13
Respuesta correcta: B. $8$
Sum $=4+7+7+9+13=40$; there are 5 values, so the mean is $40\div5=8$.
$7$ is the median and also the mode. $9$ is the range ($13-4$). $10$ divides the sum by 4.
A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random, one after the other, without replacement. What is the probability that both marbles are red?
Respuesta correcta: A. $\frac{5}{14}$
$P$(first red) = $\frac{5}{8}$. After a red marble is removed, 4 of the 7 remaining are red, so $P$(second red) = $\frac{4}{7}$. $P$(both red) = $\frac{5}{8}\frac{4}{7}=\frac{20}{56}=\frac{5}{14}$.
$\frac{25}{64}=\frac{5}{8}2$ assumes the marble is replaced. $\frac{5}{8}$ is the probability for the first draw only, and $\frac{4}{7}$ is the probability for the second draw only.
Five students, including Ana and Ben, will sit in a row of 5 chairs. In how many different orders can they sit if Ana and Ben may NOT sit next to each other?
Respuesta correcta: C. $72$
All orders: $5!=120$. Orders with Ana and Ben together: treat them as one block, giving $4!=24$ arrangements of 4 units, times 2 for the order inside the block (AB or BA): $48$. Not together: $120-48=72$.
$96$ subtracts only 24 (forgets that the pair can be AB or BA). $48$ is the number of orders where they DO sit together. $60$ is half of 120, a guess that ignores the counting.