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