Question

1. lionel tosses three fair number cubes. each number cube is labeled 1 through 6. what is the probability that all three number cubes will land on an even number? a $\frac{1}{8}$ b $\frac{1}{2}$ c $\frac{1}{6}$ d $\frac{1}{4}$

Explanation:

Step1: Probability for one die

A fair die has 6 faces (1 - 6). The even numbers are 2, 4, 6, so there are 3 even numbers. The probability of a die landing on an even number is $\frac{3}{6}=\frac{1}{2}$.

Step2: Probability for three dice

Since the tosses of the three dice are independent events, we multiply the probabilities of each event. So the probability that all three dice land on even numbers is $(\frac{1}{2})\times(\frac{1}{2})\times(\frac{1}{2})=\frac{1}{8}$. Wait, no, wait. Wait, no, let's correct. Wait, each die: even numbers are 2,4,6 (3 out of 6), so probability for one die is $\frac{3}{6}=\frac{1}{2}$. For three independent dice, the probability is $(\frac{1}{2})^3=\frac{1}{8}$? But wait, the options: A is $\frac{1}{8}$, B is $\frac{1}{2}$, C is $\frac{1}{6}$, D is $\frac{1}{4}$. Wait, but let's recheck. Wait, each die has 6 sides, even numbers are 3 (2,4,6), so probability of even on one die is $\frac{3}{6}=\frac{1}{2}$. For three dice, independent, so multiply the probabilities: $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$. So the correct answer should be A.

Answer:

A. $\frac{1}{8}$