Question

You pick a card at random. Without putting the first card back, you pick a second card at random.

5
6
7
8


What is the probability of picking an 8 and then picking an even number?

Simplify your answer and write it as a fraction or whole number.

Answer 1

To solve the problem, we need to find the probability of two events:

1. Picking an 8 on the first draw.
2. 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:

\[
\boxed{\frac{1}{12}}
\]