Four red balls and two blue balls are placed at random into two urns so that each urn contains three balls. What is the probability of getting a blue ball if
(a) You select a ball at random from the first urn?
(b) You select an urn at random and then select a ball from it at random?
(c) You discard two balls from the second urn and select the last ball?
MY ATTEMPT
Let us denote by $B$ the event "the user has selected a blue ball". There are three possible configurations for the first urn: there is no blue ball, there is one blue ball or there are two blue balls.
Let us denote by $B_{k}$ the event "there is $k$ blue balls in the first urn" where $0\leq k \leq 2$, from whence we obtain the partition $\{B_{0},B_{1},B_{2}\}$. Precisely speaking, it results that
\begin{align*}
P(B) & = P(B\cap(B_{0}\cup B_{1}\cup B_{2})) = P(B\cap B_{0}) + P(B\cap B_{1}) + P(B\cap B_{2}) =\\\\
& = 0 + P(B|B_{1})P(B_{1}) + P(B|B_{2})P(B_{2}) = \frac{1}{3}\times P(B_{1}) + \frac{2}{3}\times P(B_{2})
\end{align*}
Here is my problem: how do we calculate $P(B_{1})$ and $P(B_{2})$?
b) It suffices to notice that the selection of the urn and the selection of the ball are independent.
c) I do not know how to solve.
Am I on the right track? Any help is appreciated. Thanks in advance.
Best Answer
Fundamentally, the answer to all three parts is $\frac 13$ by symmetry. There is nothing to distinguish the ball you have picked.
You could note that $P(B_0)=P(B_2)$ because if one urn has two blues the second has none. To get $P(B_0)$ there are $20$ ways to choose three balls of six, and four of them have three red balls. So $P(B_0)=P(B_2)=\frac 15$ and $P(B_1)=\frac 35$. Then $\frac 13\cdot \frac 35+\frac 23 \cdot \frac 15=\frac 13$ as promised.