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 event

A six - sided cube has numbers 1 - 6. Even numbers are 2, 4, 6. So number of even numbers = 3. Probability of first roll being even: $P(\text{even})=\frac{3}{6}=\frac{1}{2}$.

Step2: Find probability of second event

Numbers greater than 4 on a six - sided cube are 5, 6. Number of such numbers = 2. Probability of second roll being > 4: $P(>4)=\frac{2}{6}=\frac{1}{3}$.

Step3: Find combined probability

Since the two rolls are independent events, we multiply the probabilities. So $P = \frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$. Wait, no, wait. Wait, 3 even numbers (2,4,6) so $\frac{3}{6}=\frac{1}{2}$, numbers greater than 4: 5,6 so $\frac{2}{6}=\frac{1}{3}$. Then multiply: $\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$? Wait, but let's re - check. Wait, 3 even numbers: 2,4,6 (3 out of 6), numbers greater than 4: 5,6 (2 out of 6). So the combined probability is $\frac{3}{6}\times\frac{2}{6}=\frac{6}{36}=\frac{1}{6}$? Wait, no, $\frac{3}{6}$ is $\frac{1}{2}$, $\frac{2}{6}$ is $\frac{1}{3}$, $\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$. Yes.

Answer:

$\frac{1}{6}$ (the option with $\frac{1}{6}$)