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? 1/6 1/3 2/3 5/6

Explanation:

Step1: Calculate the probability of the first roll being an even number

A six - sided 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 is \(n_1 = 3\), and the total number of outcomes is \(N=6\). The probability \(P_1=\frac{3}{6}=\frac{1}{2}\).

Step2: Calculate the probability of the second roll being a number greater than 4

The numbers greater than 4 on a six - sided cube are \(5,6\). So the number of favorable outcomes for the second roll is \(n_2 = 2\), and the total number of outcomes is \(N = 6\). The probability \(P_2=\frac{2}{6}=\frac{1}{3}\).

Step3: Use the multiplication rule for independent events

Since the two rolls are independent events, the probability \(P = P_1\times P_2\). 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:

\(\frac{1}{6}\) (the first option)