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Let $J$, $T$, and $C$ be the events that Jeff, Tim and Carmine will pass the course, respectively. | ||
Furthermore, let $G$ be the event that Professor Gore pointed out that Jeff is the one who will fail. | ||
Since the professor can't reveal Carmine's outcome, there must be $\prob{G\mid J}=0$, $\prob{G\mid T}=1$, and $\prob{G\mid C}=1/2$. | ||
Before talking to the professor, Carmine knows that his probability of passing is $\prob{C}=1/3$. | ||
His situation after the conversation is described by the event $C\mid G$. | ||
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Since $G=(G\cap J)\cup(G\cap T)\cup(G\cap C)$ and since $G\cap J$, $G\cap T$ and $G\cap C$ are mutually exclusive events, | ||
\begin{align*} | ||
\prob{G} &= \prob{G\cap J}+\prob{G\cap T}+\prob{G\cap C} \\ | ||
&= \prob{J}\prob{G\mid J}+\prob{T}\prob{G\mid T}+\prob{C}\prob{G\mid C}. | ||
\end{align*} | ||
Using Bayes's theorem, we obtain | ||
\begin{align*} | ||
\prob{C\mid G} &= \frac{\prob{C}\prob{G\mid C}}{\prob{G}} \\ | ||
&= \frac{\prob{C}\prob{G\mid C}}{\prob{J}\prob{G\mid J}+\prob{T}\prob{G\mid T}+\prob{C}\prob{G\mid C}} \\[2mm] | ||
&= \frac{(1/3)\cdot(1/2)}{(1/3)\cdot0+(1/3)\cdot1+(1/3)\cdot(1/2)} \\[1mm] | ||
&= 1/3. | ||
\end{align*} | ||
Thus, the information that Carmine has obtained does not change his chance of passing. |