Question

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

4
5
6


What is the probability of picking a prime number and then picking a prime number?

Write your answer as a percentage rounded to the nearest tenth.

%

Answer 1

To determine the probability of picking two prime numbers in succession from the cards 4, 5, and 6, we first identify the prime numbers in the set.

Among the numbers:
- 4 is not a prime number (it has divisors other than 1 and itself).
- 5 is a prime number (its only divisors are 1 and 5).
- 6 is not a prime number (it has divisors other than 1 and itself).

Thus, the only prime number present is 5.

Now, we calculate the probability of picking a prime number first and then picking a prime number again without replacement.

1. **Picking the first card**:
- The total number of cards is 3 (4, 5, 6).
- The favorable outcome (picking a prime number) is only one card (5).
- Therefore, the probability of picking a prime number first is:
\[
P(\text{first prime}) = \frac{1}{3}
\]

2. **Picking the second card** (after having picked 5):
- If we successfully pick 5 first, we have now 2 cards left (4 and 6).
- There are no prime numbers left in the remaining cards.
- Therefore, the probability of picking a prime number second is:
\[
P(\text{second prime | first prime}) = 0
\]

Now, we multiply the probabilities together to find the overall probability of both events occurring:
\[
P(\text{prime, then prime}) = P(\text{first prime}) \times P(\text{second prime | first prime}) = \frac{1}{3} \times 0 = 0
\]

Since the probability of both events occurring is zero, we convert this to a percentage:
\[
0 \times 100 = 0\%
\]

Thus, the probability of picking a prime number and then picking a prime number is:
\[
\boxed{0.0\%}
\]