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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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    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level $\endgroup$ – JKnecht Feb 4 '16 at 1:48
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    $\begingroup$ Formatting tips here. $\endgroup$ – Em. Feb 4 '16 at 1:51
  • $\begingroup$ Thank you @JKnecht and @probablyme! $\endgroup$ – Nullqwerty Feb 4 '16 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 Feb 4 '16 at 1:47
  • $\begingroup$ @Nullqwerty Almost, $1-\left(\frac{51}{52}\right)^{30} = 0.4415234 \approx 44\%$. $\endgroup$ – Em. Feb 4 '16 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 Feb 4 '16 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. Feb 4 '16 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 Feb 4 '16 at 2:18

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