# From a pack of 52

Total number of all favorable cases is n(S) = 52

Let A be the event that first card drawn is ace (or, kings). There are four aces (or, kings) in the pack. Hence, the probability of the first card is ace (or, kings) is

Let B be the event that second card is also ace (or, kings) without replacement. Then there are 3 aces (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the second card is also ace (or, kings) is

Let C be the event that third card is also ace (or, kings) without replacement. Then there are 2 aces (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the third card is also ace (or, kings) is

Let D be the event that fourth card is also ace (or, kings) without replacement. Then there are 1 ace (or, kings) left in the pack as the cards are not replaced. Therefore, the probability of the fourth card is also ace (or, kings) is

Then the probability all are aces (or, kings) is

The probability that all are aces (or, kings) is

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