To solve this problem, we need to find the probability of drawing two queens in succession without replacement from a standard deck of 52 cards.

Step-by-Step Solution:

1. First card (drawing a queen):

   • There are 4 queens in a deck of 52 cards.
   • The probability of drawing a queen on the first draw is: $P(\text{first queen}) = \frac{4}{52} = \frac{1}{13}$

2. Second card (drawing a queen after the first):

   • After drawing the first queen, there are now 3 queens left and only 51 cards remaining.
   • The probability of drawing a second queen is: $P(\text{second queen}) = \frac{3}{51} = \frac{1}{17}$

3. Total probability: Since these events are dependent (without replacement), we multiply the probabilities of the two events: $P(\text{two queens}) = P(\text{first queen}) \times P(\text{second queen}) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$

Thus, the probability of drawing two queens successively is $\frac{1}{221}$.

Would you like more details or further clarification?

Here are 5 related questions to think about:

1. What is the probability of drawing two aces successively from a deck of 52 cards?
2. How would the probability change if the cards were drawn with replacement?
3. What is the probability of drawing a queen and then a king successively?
4. What is the probability of drawing no face cards (king, queen, jack) in two draws?
5. How does the probability change if you are drawing 3 queens instead of 2?

Tip: In probability problems involving "without replacement," remember to adjust both the number of favorable outcomes and the total outcomes after each draw.