Answer

Solution

1. 1

   $P(\text{white}\mid A)=0$, $P(\text{white}\mid B)=1$, $P(\text{white}\mid C)=1/2$. Each bowl has prior $1/3$.

2. 2
   $$P(\text{white})=\tfrac{1}{3}(0+1+\tfrac{1}{2})=\tfrac{1}{2}.$$
3. 3

   We want $P(\text{other is red}\mid\text{white})$. ``Other is red'' is only possible from bowl C:

4. 4
   $$P(C\mid\text{white})=\frac{P(\text{white}\mid C)\cdot 1/3}{P(\text{white})}=\frac{(1/2)(1/3)}{1/2}=\frac{1}{3}.$$

Answer structure / marking notes

Content note