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.
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.