To solve this problem, let's first understand the setup:
1. Hiro has a stack of cards with numbers from the set [tex]\(\{1,1,2,2,3,3,3,4\}\)[/tex].
2. We need to find the probability that Hiro pulls out a 3 first and then pulls out a 2, without replacing the cards.
### Step-by-Step Solution:
1. Calculate the total number of cards:
The set of cards is [tex]\(\{1, 1, 2, 2, 3, 3, 3, 4\}\)[/tex].
Thus, the total number of cards is 8.
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]\[
P(\text{3 first}) = \frac{\text{Number of 3 cards}}{\text{Total number of cards}} = \frac{3}{8}
\][/tex]
3. Calculate the probability of pulling out a 2 second:
After pulling out a 3, there will be 7 cards left.
The remaining set of cards is [tex]\(\{1, 1, 2, 2, 3, 3, 4\}\)[/tex].
There are 2 cards with the number 2.
Thus, the probability of pulling out a 2 second is:
[tex]\[
P(\text{2 second} \mid \text{3 first}) = \frac{\text{Number of 2 cards}}{\text{Remaining number of cards}} = \frac{2}{7}
\][/tex]
4. Calculate the combined probability:
The combined probability of both events happening (pulling out a 3 first and a 2 second) is the product of the individual probabilities since these events are dependent:
[tex]\[
P(\text{3 first and 2 second}) = P(\text{3 first}) \times P(\text{2 second} \mid \text{3 first}) = \frac{3}{8} \times \frac{2}{7}
\][/tex]
5. Simplify the fraction:
[tex]\[
P(\text{3 first and 2 second}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}
\][/tex]
Therefore, the probability that Hiro pulls out a 3 first and then pulls out a 2, without replacing the cards, is:
[tex]\[
\boxed{\frac{3}{28}}
\][/tex]
