Probability of getting at least one ace |
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If this can help you approach the problem in a simpler way: you can have 4 cases: Ace, Not Ace Not Ace, Ace Ace, Ace Not Ace, Not AceTo get AT LEAST 1 ace means you have to sum the probabilities of first 3 cases. First case: `4/52*48/51` = 0.07240 Second case: is the same as the first => 0.07240 Third case: `4/52*3/51` = 0.00452Sum of the three cases is: 0.14932 Approach: At the denominator you notice that it's always 52*51 which come from the permutation formula n!/(n-r)! where n=52 and r=2. This number represents all the possible permutations of 2 cards from a deck of 52. At the numerator you need to calculate the subset of the total permutations above that should be considered as "success" cases. The first case says that you can have 4 aces in the first card and 48 'not aces' in the second. By the basic principle of counting there are 4*48 different ways to have this event. The second case follow the same reasoning approach. The third case says that you can have 4 aces in the first card and 3 left possible aces in the second card. So you have 12 possible permutations of aces.Add the first 3 success cases and divide for the overall permutations to have the probability of at least one ace. |
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