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Say we have a standard deck of $52$ cards. Probability of drawing the King of Hearts is $\frac{1}{52}$ obviously.

But lets say we were to make $30$ draws with replacement (so each time the card is drawn, it is put back in the deck and and the deck is shuffled). What are the odds that the King of Hearts card was drawn at least once out of those $30$ draws?

Thanks

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    – JKnecht
    Commented Feb 4, 2016 at 1:48
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    – Em.
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    – Nullqwerty
    Commented Feb 4, 2016 at 1:53

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Let $K$ be the number of KOHs you draw. Since each draw is with replacement and (presumably) independent of another, then it is easier to start and calculate the complement, $$P(K\geq 1) = 1-P(K=0),$$ in 30 draws.

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  • $\begingroup$ So if we filled that out, are we looking at: 1 - ((51/52)^30) ? (55.8477%) $\endgroup$
    – Nullqwerty
    Commented Feb 4, 2016 at 1:47
  • $\begingroup$ @Nullqwerty Almost, $1-\left(\frac{51}{52}\right)^{30} = 0.4415234 \approx 44\%$. $\endgroup$
    – Em.
    Commented Feb 4, 2016 at 1:51
  • $\begingroup$ That's actually really interesting. I would have thought the two were equivalent due to order of operations, but you are right, they are not. It's actually throwing me off a little since it's such a basic principle. $\endgroup$
    – Nullqwerty
    Commented Feb 4, 2016 at 1:55
  • $\begingroup$ @Nullqwerty Sure, practice helps. If you found this answer most useful or correct, consider giving it a check mark. Thanks and good luck. $\endgroup$
    – Em.
    Commented Feb 4, 2016 at 2:03
  • $\begingroup$ Oh wait sorry...duh....they are equivalent. Wow...I almost went back to school over that one. I just entered in the wrong % is all. Thought I was going crazy. Ok, thanks for the help $\endgroup$
    – Nullqwerty
    Commented Feb 4, 2016 at 2:18

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