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Two cards are drawn without replacement from an ordinary deck, find the probability that the second is a red card, given the first is a red card.

P (2nd Red Card / 1st Red Card) = 13/52 * 12 * 51 = 1/17 - is this correct?

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Your answer is not correct, no. (For starters: note that there are 26 red cards in a standard 52-card deck.)

Let $R_1$ be the event that the first card is red, and $R_2$ the event that the second card is red.

If the first card is red, then when you go to draw your second card there are a total of $26-1=25$ red cards in the deck, and a total of $52-1=51$ cards overall. So, we should have $$ P(R_2\mid R_1)=\frac{25}{51}. $$ If you don't understand this intuition, we could also go about this using the conditional probability formula: $$ P(R_2\mid R_1)=\frac{P(R_1\text{ and }R_2)}{P(R_1)}. $$ Now $$ P(R_1)=\frac{26}{52}=\frac{1}{2}, $$ since half the cards are red, and $$ P(R_1\text{ and }R_2)=\frac{26}{52}\cdot\frac{25}{51}. $$ So, we find $$ P(R_2\mid R_1)=\frac{\frac{26}{52}\cdot\frac{25}{51}}{\frac{26}{52}}=\frac{25}{51}, $$ as claimed above.

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