To determine the probability that Hiro pulls out a 3 first and then a 2 without replacing them, we can break the problem down into a series of steps.
1. Determine the total number of cards:
The set of cards is given as [tex]\( \{1, 1, 2, 2, 3, 3, 3, 4\} \)[/tex].
There are 8 cards in total.
2. Calculate the probability of pulling out a 3 first:
There are 3 cards with the number 3.
Therefore, the probability of pulling out a 3 first is:
[tex]\[
\text{P(pulling a 3 first)} = \frac{\text{Number of 3's}}{\text{Total number of cards}} = \frac{3}{8}
\][/tex]
3. Determine the new total number of cards after one has been pulled:
Once a card is pulled out, there are 7 cards left.
4. Calculate the probability of pulling out a 2 next:
After pulling out one 3, there are still 2 cards with the number 2 in the remaining 7 cards.
Therefore, the probability of pulling out a 2 next is:
[tex]\[
\text{P(pulling a 2 second)} = \frac{\text{Number of 2's left}}{\text{Total number of cards left}} = \frac{2}{7}
\][/tex]
5. Multiply these probabilities to get the final answer:
To get the overall probability of first drawing a 3 and then drawing a 2:
[tex]\[
\text{Total probability} = \text{P(pulling a 3 first)} \times \text{P(pulling a 2 second)} = \frac{3}{8} \times \frac{2}{7} = \frac{3 \cdot 2}{8 \cdot 7} = \frac{6}{56}
\][/tex]
6. Simplify the fraction if possible:
[tex]\[
\frac{6}{56} = \frac{3}{28}
\][/tex]
Therefore, the probability that Hiro pulls out a 3 first and then a 2 without replacing them is:
[tex]\[
\boxed{\frac{3}{28}}
\][/tex]
