#1502 C.2-8

appendixc/sections/2/8.tex

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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 $C\mid G$.
+
+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.