Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\frac{13}{52})(\frac{13}{51})(\frac{2}{50})$ + $2(\frac{13}{52})(\frac{13}{51})(\frac{3}{50})$ + $(\frac{13}{52})(\frac{13}{51})(\frac{4}{50})$
= $\frac{13}{850}$
Which answer is correct?

$$P(A) = \frac{n(A)}{n(\Omega)}$$
Here, $$n(A) = n(\text{ace of spade}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{spade other than ace}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{ace of spade}) \times n(\text{ heart other than ace}) \times n(\text{other aces}) + \\ n(\text{spade other than ace}) \times n(\text{ heart other than ace}) \times n(\text{aces})$$
$$=1 \times 1 \times 2+ 2 \times 12 \times 1 \times 3 + 12 \times 12 \times 4$$
$$ = 2+72+576=650$$
$$n(\Omega)= 52 \times 51 \times 50 $$
So, $P(A) =1/204$.
