Bag of colored tiles. Probability that the 3 drawn are different Tiles in bag: 10 Black, 1 Blue, 1 Red, 1 Green, 1 Yellow, 1 Orange
I draw three at random and tell you that one of the ones  I drew is red. What is the probability that all three of the tiles I drew are different? 
I tried setting this up as a Baye's Theorem question, but I got stuck.
Pr(All different colors|One is red) is what I'm solving for.
To get the probability that all are a different color, I found the probability of combinations with 2 or 3 of the same color. Then I subtracted it from 1. 
Pr(All different) = .2418
Pr(Not all different) = .758
Pr(One is red|All different) = .2757
Pr(One is red|Not all different) = .0879
I put this in Baye's formula: (.2418*.2757)/((.2418*.2757)+(.758*.0879)) and got ~.50 
I HAVE NO IDEA IF THIS IS RIGHT. I think I messed up along the way. 
I am stuck. 
 A: The answer is close to right, you did some rounding. 
Let $R$ be the event "one of the tiles is red" and $D$ the event "the colours are are all different." We want $\Pr(D|R)=\frac{\Pr(R\cap D)}{\Pr(R)}$.
There are $\binom{15}{3}$ equally likely ways to draw $3$ tiles. (We treat the black tiles as having student numbers). 
To find $\Pr(R)$, we find the number of ways to have $1$ red. We can either have $2$ black, giving $\binom{10}{2}$ choices, or $1$ black and $1$ non-red non-black, giving $\binom{10}{1}\binom{4}{1}$ choices, or $2$ non-red non-black, giving $\binom{4}{2}$ choices, a total of $91$. Thus
$$\Pr(R)=\frac{91}{\binom{15}{3}}.$$
A similar calculation shows that the number of ways to have  $1$ red and  all colours different is $\binom{10}{1}\binom{4}{1}+\binom{4}{2}$, and therefore
$$\Pr(R\cap D)=\frac{46}{\binom{15}{3}}.$$
Finally, divide. 
A: There's really no need to use Bayes' theorem on a problem like this (unless the assignment requires it).  As soon as you know one of the tiles is red, the probability requested is simply the probability that the other two tiles, which can be thought of as drawn at random from a bag now containing just $14$ tiles, aren't both black.  That is,
$$1-{{10\choose2}\over{14\choose2}}=1-{10\cdot9\over14\cdot13}={182-90\over182}={46\over91}$$
