Probability of taking two red balls from urn I have a very basic question: if I have an urn with 5 black balls and 11 red balls, I want to know what the probability is that I pick two red balls.
The way I'm currently thinking about this is with simple conditional probability:
p(b | a) = p(a and b)/p(a)

I know that given one pick, the probability of choosing a red ball is 11/16. So, I assume that p(red ball | first ball red) = p(a)p(b)/p(a) = (11/16)(10/15)/(11/16)
Something about this seems incorrect to me though. Can I go from p(red ball | first ball red) to p(a)p(b)? Also, is the p(red ball | first ball red) to p(a)p(b) even the probability of choosing two red balls?
 A: On your first pick, the probability is 11/16 that you choose a red ball.  On the second pick, the probability is 10/15 that you choose a red ball.  So 11/16*10/15 = 110/240 = 45.83% chance of picking two red balls in a row.  (I think - someone please correct me if I'm wrong). 
A: Lisa Williams has correctly calculated the probability.  As she observed, the probability of selecting a red ball on the first draw is $11/16$ since $11$ of the original $16$ balls are red.  Since we are drawing without replacement, the probability of selecting a second red ball is $10/15$ since $10$ of the $15$ remaining balls are red.  Thus, the probability of selecting two red balls without replacement is 
$$P(\text{two red}) = P(\text{red})P(\text{red} \mid \text{first ball is red}) = \frac{11}{16} \cdot \frac{10}{15} = \frac{11}{24}$$
Observe that the probability that the second ball selected is red depends on the fact that the first ball was red.  
The condition that 
$$P(F \mid E) = \frac{P(E \cap F)}{P(E)} = \frac{P(E)P(F)}{P(E)} = P(F)$$
means that events $E$ and $F$ are independent since the probability that event $F$ occurs given that event $E$ has occurred is equal to the probability that event $F$ occurs.  That is not the case here since the probability that the second ball selected is red depends on whether the first ball selected was red, so $P(\text{red} \mid \text{first ball is red}) \neq P(\text{red})P(\text{red})$.  It would be the case if the balls were replaced after each draw.  
An alternative method of calculating the probability of selecting two red balls without replacement is 
$$P(\text{both red}) = \frac{\binom{11}{2}}{\binom{16}{2}}$$
where the numerator represents the number of ways of selecting two of the eleven red balls in the urn and the denominator represents the number of ways of selecting two of the sixteen balls in the urn. 
