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Say we have a standard deck of 52 cards. Probability of drawing the King of Hearts OR the Queen of Diamonds is $\frac{2}{52}$ obviously (or $\frac{1}{26}$). If 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), the odds that the King of Hearts OR the Queen of Diamonds card was drawn at least once out of those 30 draws? And the answer is $1-\left((\frac{50}{52}\right)^{30})$

But now, what if we had the same situation but instead had asked: What are the odds that the King of Hearts OR the Queen of Diamonds card was drawn at least ten times out of those 30 draws?

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Draws are with replacement. Notice that getting a KOH or QOD per trial in 30 trials follows a binomial distribution. So let $X$ be the number of times this happens; then $$P(X\geq 10) = \sum_{k=10}^{30}\binom{30}{k}\left(\frac{2}{52}\right)^k\left(\frac{50}{52}\right)^{30-k} = 1-\sum_{k=0}^9\binom{30}{k}\left(\frac{2}{52}\right)^k\left(\frac{50}{52}\right)^{30-k}$$

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  • $\begingroup$ Aren't you missing a binomial coefficient in the last expression? $\endgroup$
    – Brian Tung
    Commented Feb 4, 2016 at 22:51
  • $\begingroup$ @BrianTung Typo, thanks! $\endgroup$
    – Em.
    Commented Feb 4, 2016 at 22:54
  • $\begingroup$ That's awesome! Thank you again! Super helpful! $\endgroup$
    – Nullqwerty
    Commented Feb 4, 2016 at 22:55

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