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 (even)

A six - sided cube has numbers 1 - 6. Even numbers: 2, 4, 6 (3 numbers). Probability \( P(\text{even})=\frac{3}{6}=\frac{1}{2} \).

Step2: Find probability of second roll (>4)

Numbers > 4: 5, 6 (2 numbers). Probability \( P(>4)=\frac{2}{6}=\frac{1}{3} \).

Step3: Multiply the two probabilities

Since the two rolls are independent, \( P = \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} \)? Wait, no, wait. Wait, no, let's re - check. Wait, first roll even: numbers 2,4,6: 3 out of 6, so \( \frac{3}{6}=\frac{1}{2} \). Second roll greater than 4: numbers 5,6: 2 out of 6, so \( \frac{2}{6}=\frac{1}{3} \). Then the combined probability is \( \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} \)? But wait, the options have \( \frac{1}{6} \) as an option. Wait, but let's check again. Wait, maybe I made a mistake. Wait, first roll: even numbers (2,4,6) → 3 numbers. So \( P(\text{even})=\frac{3}{6}=\frac{1}{2} \). Second roll: numbers greater than 4 are 5 and 6, so 2 numbers. \( P(>4)=\frac{2}{6}=\frac{1}{3} \). Multiply them: \( \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} \). Wait, but the options have \( \frac{1}{6} \) as the first option. Wait, but let's check the calculation again. Wait, no, wait, maybe I messed up the second probability. Wait, numbers greater than 4 on a six - sided die: 5 and 6, so 2 numbers. So \( \frac{2}{6}=\frac{1}{3} \). First roll even: 3 numbers, \( \frac{3}{6}=\frac{1}{2} \). Then the probability of both events is \( \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} \).

Answer:

\(\frac{1}{6}\) (the first option, i.e., the option with \(\frac{1}{6}\))