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.
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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: \[ P(\text{first prime}) = \frac{1}{3} \]
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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: \[ \boxed{0.0%} \]