Question

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

1
2
3
4
5
6


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

Write your answer as a percentage.

%

Answers

GPT-4o mini
To solve the problem, we first identify the odd and even numbers in the set of cards:

The cards are: 1, 2, 3, 4, 5, 6.

The odd numbers in this set are: 1, 3, 5.

So there are 3 odd numbers out of a total of 6 cards.

**Step 1: Probability of picking an odd number first**

The probability of picking an odd number on the first draw is:

\[
P(\text{odd first}) = \frac{\text{Number of odd numbers}}{\text{Total number of cards}} = \frac{3}{6} = \frac{1}{2}
\]

**Step 2: Probability of picking an odd number second, given the first was odd**

If the first card drawn is odd, there will be now 2 odd cards left and 5 cards total left (since we do not put the card back).

The probability of picking an odd number on the second draw is:

\[
P(\text{odd second | odd first}) = \frac{\text{Remaining odd numbers}}{\text{Total remaining cards}} = \frac{2}{5}
\]

**Step 3: Total probability of both events occurring**

To find the total probability of both events (picking an odd number first and then picking another odd number), we multiply the probabilities of the two independent events:

\[
P(\text{odd first and odd second}) = P(\text{odd first}) \times P(\text{odd second | odd first}) = \frac{1}{2} \times \frac{2}{5} = \frac{2}{10} = \frac{1}{5}
\]

**Step 4: Convert to percentage**

To convert the probability to a percentage, we multiply by 100:

\[
\frac{1}{5} \times 100 = 20\%
\]

Thus, the probability of picking an odd number and then picking an odd number is \(\boxed{20\%}\).