Probability with balls I keep playing with this and get different answers - help!
Urn contains 4 black balls and 5 white.  A ball is drawn at random and then replaced after which the second ball is drawn.  What is the probability that the first is black and the second is white?
 A: Hint: use the product rule.
$P(b)$: probability of drawing a black ball.
$P(w)$ : probability of drawing a white ball.
Then we have that the probability of one, followed by the other, with replacement, is $$P(b)\times P(w)$$
I'll let you think about what the value of $P(b)$ and of $P(w)$, for which you have all the information you need:
$P(b) = \dfrac{\text{# of black balls}}{\text{total # of balls}}$
$P(w) =  \dfrac{\text{# of white balls}}{\text{total # of balls}}$
A: The probability the first is black is $\dfrac{4}{9}$. Given that the first is black, the probability the second is white is $\dfrac{5}{9}$. The desired probability is the product of these.
Alternately, imagine that the balls have ID numbers on them. We record the result of the draws as an ordered pair of ID numbers. There are $9^2$ such ordered pairs, all equally likely.
We now count the favourables, the ordered pairs that consist of one of the black balls, followed by one of the white. There are $(4)(5)$ such ordered pairs, Thus our probability is $\dfrac{(4)(5)}{9^2}$. 
