Two cards are drawn without replacement. Find a probability when Queens are drawn.

Question

Standard deck of 52 cards. 2 cards are dealt WITHOUT replacement. What is the probability that both cards are Queens given that one of them is the Queen of Clubs.

Conditional Probability.

P(Two Queens | Queen of Clubs) = P(Two Queens AND Queen of Clubs) / P(Queen of Clubs)

P(Two cards of the same face value) = (4/52)(3/51) = 1/221

Stuck.

Answer 1

There are $51$ two-card hands that contain the queen of clubs. Of these, $3$ contain another queen. Hence, the desired probability is

$$\frac{3}{51} = \frac{1}{17}$$