In a 2-card hand, what is the probability of holding 2 kings?

A pack of cards has 52 cards, 4 of which are kings, so the probability of the first card being a king is (4/52). Since you have already chosen a card, there are only 51 cards left to choose from in the entire deck (the first card is not counted anymore because it has already been pulled from the deck), and since the card you chose was a king, you are left with only 3 more kings, meaning that the probability of the second card being a king is (3/51). Since we are interested in both happening simultaneously, we must multiply the probability of the first by the probability of the second: (4/52) • (3/51), which leaves us with: * 12/2,652 or 0.0045249 *

what "following statement"?

step-by-step explanation:

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: