# Math

Drawing a least one king when you draw a card from a standard deck 8 times (replacing the card each time you draw, so there are always 52 cards in the deck)

P(at least one king in 8 attempts)= [P(1 king + 2 kings + 3 kings + 4 kings]^8 =
1/52 + 2/52 + 3/52 + 4/52 = 10/52^8

Not sure how to solve, should I subtract 1 from the answer.

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1. prob(a king) = 4/52 = 1/13
prob(not a king) = 12/13

let's find the prob of NOT getting a king, 8 times in a row
= (12/13)^8
but we DON'T want that to happen

so prob of at least one king somewhere
= 1 - (12/13)^8
= appr .4729

btw, your attempt makes no sense to me

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