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 probability of first - roll being even

The sample space of rolling a six - sided die is \(S = \{1,2,3,4,5,6\}\). The even numbers are \(E=\{2,4,6\}\). The probability of getting an even number on the first roll, \(P(A)=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\).

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

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

Step3: Use the multiplication rule for independent events

Since the two rolls of the die are independent events, the probability that the first roll is an even number and the second roll is a number greater than 4 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: \(P(A\cap B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\).

Answer:

\(\frac{1}{6}\)