Q.Urn $A$ contains $5$ white and $3$ black balls, Urn $B$ contains $3$ white and $5$ black balls, and Urn $C$ contains $4$ white and $4$ black balls. An urn is chosen at random and a ball is drawn from it. Find the probability that the ball drawn is white.

[!TLDR] Each urn equally likely ($\tfrac13$); total probability over $A,B,C$. [!ANSWER] $P(W)=\dfrac12$.

$P(A)=P(B)=P(C)=\tfrac13$. $P(W\mid A)=\tfrac58$, $P(W\mid B)=\tfrac38$, $P(W\mid C)=\tfrac48$. By the total probability theorem, $P(W)=\tfrac13\left(\tfrac58+\tfrac38+\tfrac48\right)=\tfrac13\times\tfrac{12}{8}=\tfrac13\times\tfrac32=\tfrac12$. [!ANSWER] $P(W)=\dfrac12$.

Weight each urn's conditional probability of white by $\tfrac13$ (equal chance of selecting each urn) and sum.

Averaging the three conditional probabilities by adding the fractions with different-looking numerators without a common denominator first is a frequent slip.