Question

Two six-sided number cubes are rolled. Which event is more likely?
Event 1: rolling two even numbers
Event 2: rolling two sixes(1 point)
Event

Answer 1

To determine which event is more likely when rolling two six-sided number cubes (dice), we can calculate the probability of each event.

**Event 1: Rolling two even numbers.**
The even numbers on a six-sided die are 2, 4, and 6. Therefore, there are 3 even numbers out of 6 possible outcomes.

- The probability of rolling an even number on the first die is \(\frac{3}{6} = \frac{1}{2}\).
- The probability of rolling an even number on the second die is also \(\frac{3}{6} = \frac{1}{2}\).

Since the rolls are independent, the probability of both dice showing even numbers is:
\[
P(\text{both even}) = P(\text{even on die 1}) \times P(\text{even on die 2}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.
\]

**Event 2: Rolling two sixes.**
There is only one outcome for each die that results in a six.

- The probability of rolling a six on the first die is \(\frac{1}{6}\).
- The probability of rolling a six on the second die is also \(\frac{1}{6}\).

The probability of both dice showing sixes is:
\[
P(\text{both sixes}) = P(\text{six on die 1}) \times P(\text{six on die 2}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}.
\]

**Comparison of probabilities:**
- Probability of Event 1 (two even numbers) = \(\frac{1}{4} = \frac{9}{36}\) (when expressed with a common denominator).
- Probability of Event 2 (two sixes) = \(\frac{1}{36}\).

Since \(\frac{9}{36} > \frac{1}{36}\), Event 1 (rolling two even numbers) is more likely than Event 2 (rolling two sixes).