Question

The table below shows the possible outcomes of rolling a six-sided number cube and flipping a coin.

A 7-column table with 2 rows. Column 1 has entries H, T. Column 2 is labeled 1 with entries H 1, T 1. Column 3 is labeled 2 with entries H 2, T 2. Column 4 is labeled 3 with entries H 3, T 3. Column 5 is labeled 4 with entries H 4, T 4. Column 6 is labeled 5 with entries H 5, T 5. Column 7 is labeled 6 with entries H 6, T 6.

What is the probability of getting a number less than 3 and a tails?
StartFraction 1 over 12 EndFraction
StartFraction 1 over 6 EndFraction
One-fourth
One-third

Answer 1

To determine the probability of getting a number less than 3 and a tails when rolling a six-sided die and flipping a coin, we can break down the problem as follows:

1. **Outcomes of the die:**
- The possible outcomes of rolling a six-sided die are 1, 2, 3, 4, 5, and 6.
- The outcomes that are less than 3 are: 1 and 2. So, there are 2 favorable outcomes for the die.

2. **Outcomes of the coin:**
- The coin can result in either heads (H) or tails (T). So there is 1 favorable outcome for getting tails.

3. **Total combinations:**
- The total number of outcomes when rolling the die and flipping the coin is:
- 6 outcomes from the die multiplied by 2 outcomes from the coin.
- Total outcomes = 6 × 2 = 12.

4. **Favorable outcomes for the event (number < 3 and tails):**
- The favorable outcomes that meet our criteria (number less than 3 and tails) are:
- H 1, T 1
- H 2, T 2
- The two possibilities with Tails are:
- T 1
- T 2
- Therefore, there are 2 favorable outcomes: (T 1) and (T 2).

5. **Calculating probability:**
- The probability is calculated as the number of favorable outcomes divided by the total number of outcomes:
\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{12} = \frac{1}{6}.
\]

Based on this calculation, the probability of getting a number less than 3 and a tails is \( \frac{1}{6} \).

So, the correct answer is \(\frac{1}{6}\).