Two cards are drawn without replacement from an ordinary deck, find the probability..

Question

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?

Answer 1

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.

Answer 2

Can the alternate form of conditional probability be used here i.e.

$$P(R_2 | R_1) = \frac{n(R_1 \bigcap R_2)}{n(R_1)}$$

We've promoted this method since the product rule hasn't been taught yet.

thanks