I am working on this question that involves a player having 2 cards in their hand, an Ace of clubs and a King of clubs. Out of the 50 remaining cards, the player will be given another 5 cards at random for a total of 7 in their hand. What is the probability that at least one of the 5 new cards are either an ace or a king?
My problem is that I don't understand how to mathematically apply the "at least" statement. I know that since the player already has an ace and a king in their hand that the probability of getting either an ace or a king on the next draw is 6/50, then it becomes 6/49 for the next if they still fail to draw an ace or a king and so on and so forth. However, I think my logic is flawed and I am stumped as of now. Any help would be appreciated.
Ask it the other way around. What is the probability of not drawing any kings or aces? That is $\frac{44×43×42×41×40}{50×49×48×47×46} = 0.51$. You seek the probability of this not hapening, or $1 - 0.51 = 0.49$