Question

a six - sided number cube is rolled twice. what is the probability that the first roll is an even number and the second roll is a number greater than 4?
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{5}{6}$

Explanation:

Step1: Find probability of first - roll being even

A six - sided number cube has numbers 1, 2, 3, 4, 5, 6. The even numbers are 2, 4, 6. So the number of favorable outcomes for the first roll being even is 3, and the total number of outcomes is 6. The probability $P_1$ of the first roll being even is $P_1=\frac{3}{6}=\frac{1}{2}$.

Step2: Find probability of second - roll being greater than 4

The numbers greater than 4 on a six - sided number cube are 5 and 6. So the number of favorable outcomes for the second roll being greater than 4 is 2, and the total number of outcomes is 6. The probability $P_2$ of the second roll being greater than 4 is $P_2 = \frac{2}{6}=\frac{1}{3}$.

Step3: Use the multiplication rule for independent events

Since the two rolls are independent events, the probability $P$ that the first roll is an even number and the second roll is a number greater than 4 is $P=P_1\times P_2$. Substitute $P_1=\frac{1}{2}$ and $P_2=\frac{1}{3}$ into the formula: $P=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$