Question

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

6
7
8


What is the probability of picking a 6 and then picking an odd number?

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

Answer 1

To solve for the probability of picking a 6 and then an odd number from the given set of cards, we start by determining the total number of cards.

The cards are:
- 6
- 7
- 8

There are 3 cards total.

### Step 1: Probability of picking a 6
The probability of picking a 6 as the first card is given by:
\[
P(\text{picking a 6}) = \frac{1}{3}
\]

### Step 2: Probability of picking an odd number after picking a 6
After picking the 6, the remaining cards are:
- 7
- 8

From these, the odd number is 7. There are 2 cards remaining (7 and 8), and only one of them (7) is an odd number.

Thus, the probability of picking an odd number after picking a 6 is:
\[
P(\text{picking an odd number | picked a 6}) = \frac{1}{2}
\]

### Step 3: Combining the probabilities
Now, we multiply the probabilities of the two independent events (picking a 6 first and then picking an odd number):
\[
P(\text{6 and then an odd number}) = P(\text{picking a 6}) \times P(\text{picking an odd number | picked a 6})
\]
\[
P(\text{6 and then an odd number}) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}
\]

Thus, the probability of picking a 6 and then picking an odd number is:
\[
\boxed{\frac{1}{6}}
\]