Question

Daniel is playing a game with his friend where he draws a tile and a card from a box filled with four different colored square tiles and three cards numbered 1, 2, and 3. The table below represents all the possible outcomes when drawing a tile and a card.

Outcomes Red Blue Yellow Green
1 1, Red 1, Blue 1, Yellow 1, Green
2 2, Red 2, Blue 2, Yellow 2, Green
3 3, Red 3, Blue 3, Yellow 3, Green

What is the probability of Daniel drawing a card numbered 1 or 2 and a yellow tile?
A.

B.

C.

D.

Answer 1

To determine the probability of Daniel drawing a card numbered 1 or 2 and a yellow tile, we first need to identify the total number of possible outcomes and then the number of favorable outcomes.

From the table, we can see the outcomes:

1. **Total Outcomes**: There are 12 total outcomes since there are 3 cards (1, 2, 3) and 4 colored tiles (Red, Blue, Yellow, Green):

- Card 1: 1, Red; 1, Blue; 1, Yellow; 1, Green (4 outcomes)
- Card 2: 2, Red; 2, Blue; 2, Yellow; 2, Green (4 outcomes)
- Card 3: 3, Red; 3, Blue; 3, Yellow; 3, Green (4 outcomes)

Therefore, there are \( 3 \text{ (cards)} \times 4 \text{ (tiles)} = 12 \text{ (total outcomes)} \).

2. **Favorable Outcomes**: We need to count the outcomes that consist of either card 1 or card 2 **and** a yellow tile:

- For card 1: 1, Yellow (1 outcome)
- For card 2: 2, Yellow (1 outcome)

This gives us a total of \( 1 + 1 = 2 \) favorable outcomes.

3. **Calculating Probability**: Probability is calculated using the formula:

\[
P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\]

Plugging in our values:

\[
P(\text{drawing card 1 or 2 and yellow tile}) = \frac{2}{12} = \frac{1}{6}
\]

Thus, the final answer is:

**Probability** = \(\frac{1}{6}\)

If options A, B, C, D were to represent numerical values (e.g., \(\frac{1}{6}\)), you should select the one corresponding to this probability.