I have a pack of cards. I draw one card,then a second card without putting the first card back in the pack. What is the probability that I draw two aces?
prob(2aces) = (4/52)(3/51) = ...
To calculate the probability of drawing two aces, we need to consider the total number of possible outcomes and the number of favorable outcomes.
First, let's determine the total number of possible outcomes. In a standard deck of 52 cards, there are 4 aces. Since we draw one card and then draw a second card without replacement, the total number of possible outcomes for the first card is 52, and for the second card, it is 51.
Now let's calculate the number of favorable outcomes, which are the two aces together. We have 4 aces in the deck, and once we draw one ace, there are 3 aces remaining. So, the number of favorable outcomes is 4 * 3 = 12.
Therefore, the probability of drawing two aces is given by the ratio of favorable outcomes to total outcomes:
P(Two aces) = Favorable outcomes / Total outcomes
P(Two aces) = 12 / (52 * 51)
P(Two aces) ≈ 0.0045
So, the probability of drawing two aces is approximately 0.0045, or about 0.45%.