To determine the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacing the cards, let's consider the scenario step-by-step.
1. Determine the total number of cards and their counts:
- Total number of cards = 8.
- Cards: [tex]\(1, 1, 2, 2, 3, 3, 3, 4\)[/tex]
- Number of 3's = 3.
- Number of 2's = 2.
2. Calculate the probability of pulling out a 3 first:
- The number of favorable outcomes for drawing a 3 is 3.
- Therefore, the probability of drawing a 3 first is:
[tex]\[
\frac{\text{Number of 3's}}{\text{Total number of cards}} = \frac{3}{8}
\][/tex]
3. Adjust the total number of cards after pulling out the first card:
- After drawing a 3, the total number of cards is reduced by 1, so now there are 7 cards left.
4. Calculate the probability of pulling out a 2 next:
- After one 3 is removed, the remaining cards are: [tex]\(1, 1, 2, 2, 3, 3, 4\)[/tex]
- The number of favorable outcomes for drawing a 2 now is 2.
- Therefore, the probability of drawing a 2 after already having drawn a 3 is:
[tex]\[
\frac{\text{Number of 2's left}}{\text{Total remaining cards}} = \frac{2}{7}
\][/tex]
5. Combine the two independent probabilities:
- To find the combined probability that Hiro pulls out a 3 first and then pulls out a 2 (in this order), we multiply the individual probabilities:
[tex]\[
\left( \frac{3}{8} \right) \times \left( \frac{2}{7} \right) = \frac{3 \times 2}{8 \times 7} = \frac{6}{56} = \frac{3}{28}
\][/tex]
Therefore, the probability that Hiro pulls out a 3 first and then pulls out a 2 without replacing them is:
[tex]\[
\boxed{\frac{3}{28}}
\][/tex]
