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Suppose the two joker cards are left in a standard deck of cards. One joker is red and the other is black. A single card is drawn from the deck of 54 cards, returned to the deck, and a second card is drawn. Determine the probability of drawing a red joker or a red ace on either draw.

So there are two red aces and one red joker.

Instinctively, I would think $\frac{3}{54}\times\frac{3}{54}=\frac{1}{324} $

But 1 - Pr(no red joker or red ace on either draw) = $1-\frac{51}{54}\times\frac{51}{54}=\frac{35}{324} $

I can't figure this out...

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    $\begingroup$ $\frac{3}{54}\times\frac{3}{54}$ is the probability that both draws are a red Joker or a red Ace. $1-\frac{51}{54}\times\frac{51}{54}$ that at least one draw is a red Joker or a red Ace. $\endgroup$
    – Henry
    Commented Mar 21, 2022 at 14:37
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    $\begingroup$ Quite... If you insist on approaching directly, then this would be "both first and second draw" + "first but not second" + "second but not first" leading to a calculation of $\frac{3}{54}\times\frac{3}{54}+\frac{3}{54}\times\frac{51}{54}+\frac{51}{54}\times\frac{3}{54}$, or perhaps a bit cleaner as "first" + "second but not first" leading to $\frac{3}{54}+\frac{51}{54}\times\frac{3}{54}$, both of which of course equal $\frac{35}{324}$, same as the indirect method yields. $\endgroup$
    – JMoravitz
    Commented Mar 21, 2022 at 14:40

1 Answer 1

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The question asks you to determine the probability of drawing a red joker or a red ace on either draw. So your second attempt using complimentary method is correct. In your first approach you are calculating the probability that both draws are either a red ace or a red joker. The correct answer would be,

$ \displaystyle \frac{3}{54} \cdot \frac{51}{54} + \frac{51}{54} \cdot \frac{3}{54} + \frac{3}{54} \cdot \frac{3}{54} = \frac{35}{324}$

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