From a standard 52 card deck, you draw two cards without replacement. What is the probability that both cards are red knowing at least 1 is red?

Question

The probability of a second drawn card being red is (26-1)/(52-1) = (25/51) because we already know one of the 26 red cards is drawn and one is removed from the complete deck.

Is there anything I am missing to account for?

Answer 1

Let the event A is "both cards are red", the event B is "at least 1 is red".

P(A | B) = P(A ∩ B) / P(B)

A ∩ B = "both cards are red and at least 1 is red" = "both cards are red" = A

So, P(A ∩ B) = P(both cards are red) = (26/52)(25/51) = 25/102

P(B) = P(at least 1 is red) = 1 - P(both cards are black) = 1 - (26/52)(25/51) = 77/102

P(both cards are red | at least 1 is red) = (25/102) / (77/102) = 25/77