Question:
A person draws two card and shows one of them, which is an Ace. What is the probability that other card is also an Ace?
My Approaches:
Since it is given that one card is Ace, there are 51 cards left. 3 of them are Aces. Probability is simply 3/51.
SInce the question is what is the probability of another ace it should be (Probability of one Ace)*(Probability of 2nd Ace) i.e. (4/52)(3/51)
Counter Arguments: My friend thinks it is not right. And propose following:
1st approach is not correct. It simply implies choosing of one Ace from a decck from which an Ace is removed.
2nd approach is not correct as it implies probability of choosing of two aces one after another. Does not really fits to the case that given a person already drawn an ace and now talking about only the second card.
Any help will be appreciated.