To solve the problem, we need to find the probability of two events:
- Picking an 8 on the first draw.
- Picking an even number (which are 6 and 8 from the given cards) on the second draw.
Let's start by identifying our set of cards, which are: 5, 6, 7, 8.
Step 1: Probability of picking an 8 first
There are 4 cards in total. The probability \( P(\text{8 first}) \) of picking an 8 on the first draw is:
\[ P(\text{8 first}) = \frac{1 \text{ (number of 8s)}}{4 \text{ (total number of cards)}} = \frac{1}{4} \]
Step 2: Probability of picking an even number second
After picking the 8, we have 3 cards left, which are: 5, 6, and 7.
Out of these 3 cards, only one card is an even number (which is 6). The probability \( P(\text{even second} | \text{8 first}) \) of picking an even number (6) on the second draw, given that we picked 8 first, is:
\[ P(\text{even second} | \text{8 first}) = \frac{1 \text{ (number of evens remaining)}}{3 \text{ (remaining cards)}} = \frac{1}{3} \]
Step 3: Combined Probability
Now, we can find the combined probability of both events occurring in sequence. This is done by multiplying the probabilities of the two events:
\[ P(\text{8 first and even second}) = P(\text{8 first}) \times P(\text{even second} | \text{8 first}) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \]
Final Answer
The probability of picking an 8 first and then picking an even number is:
\[ \boxed{\frac{1}{12}} \]