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?

Explanation:

Step1: Calculate first - roll probability

A standard six - sided die has 6 possible outcomes: 1, 2, 3, 4, 5, 6. The even numbers are 2, 4, 6. So the probability of rolling an even number on the first roll, $P(A)=\frac{3}{6}=\frac{1}{2}$.

Step2: Calculate second - roll probability

The numbers greater than 4 on a six - sided die are 5 and 6. So the probability of rolling a number greater than 4 on the second roll, $P(B)=\frac{2}{6}=\frac{1}{3}$.

Step3: Use multiplication rule for independent events

Since the two rolls of the die are independent events, the probability of both events occurring is $P(A\cap B)=P(A)\times P(B)$. Substitute $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$ into the formula, we get $P(A\cap B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$