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\newcommand{\mytitle}[1]{\begin{center} {\Large Intro to Cognitive Science -- 
9.62J Problem Set \psnum \\ Juliet Bourne -- #1} \\ \vspace{-.1in}
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{\Large 9.62J Problem Set \psnum -- Juliet Bourne -- {\bf Page \thepg}}
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\begin{document}

\mytitle{11/15/90}

\vspace*{-.15in}
\begin{center} {\large Probability Judgment} \end{center}

\vspace*{-.2in}

\prob{3.1} 

\subprob{a} Let $R$ stand for the statement that the Red Sox win the AL
pennant. Let $C$ stand for the statement that the Celtics win the NBA
championships. 

\vspace{.1in}
\beq
P_{Jones}(R \vee C) & = & \frac{5}{8} \\
P_{Jones}(\neg R) = \frac{4}{5} & \Rightarrow & P_{Jones}(R) = \frac{1}{5} \\
P_{Jones}(\neg C) = \frac{3}{4} & \Rightarrow & P_{Jones}(C) = \frac{1}{4} \\ 
\eeq

$P_{Jones}$ is not Bayesian, since if it were, by condition (7c):

\beq 
P_{Jones}(R \vee C) & = & P_{Jones}(R) + P_{Jones}(C) \\
\mbox{} & = & \frac{1}{5} + \frac{1}{4} \\
\mbox{} & = & \frac{9}{20}
\eeq

Thus, since $P_{Jones}$ is not Bayesian and since Jones is willing to
accept moderate fair bets, Jones is open to a Dutch book.

\subprob{b} Let $L$ stand for the statement ``Linda is a bank teller''.
Let $F$ stand for the statement ``Linda is active in the feminist
movement''. 

\beq 
P_{me}(L) = 0.1 & \; \; \; P_{me}(L \wedge F) = 0.2 
\eeq

These two probabilities show that $P_{me}$ is not Bayesian since
condition (8d) says: $P_{me}(L \wedge F) \le P_{me}(L)$. But in my case,
$P_{me}(L \wedge F) = 0.2$ which is greater than $P_{me}(L) = 0.1$.
Therefore, since $P_{me}$ is not Bayesian, I am open to a Dutch book.

\prob{3.2}

\beq
P(S_1 | S_2) & = & 1 - P(\neg S_2 | S_1) \\
\frac{P(S_1 | S_2) P(S_2)}{P(S_1)} & = & 1 - \frac{P(S_1 | \neg S_2)
P(\neg S_2)}{P(S_1)} \\
1 & = & \frac{P(S_1 | S_2) P(S_2)}{P(S_1)} + \frac{P(S_1 | \neg S_2)
P(\neg S_2)}{P(S_1)} \\
P(S_1) & = & P(S_1 | S_2) P(S_2) + P(S_1 | \neg S_2) P(\neg S_2)
\eeq

\prob{3.3} Let $FP$ indicate that the result of the test was a false
positive. Let $D$ indicate that the person being tested actually has the
disease. 

\beq
P_{test} & = & 0.05 \\
P_{test} & = & 0.001 \\
P_{test}(\neg FP | D) & = & 1 \\
P_{test}(D | \neg FP) & = & \frac{P_{test}(\neg FP | D)
P_{test}(D)}{P_{test}(\neg FP)} \\
\mbox{} & = & \frac{(1)(0.001)}{0.05} \\
P_{test}(D | \neg FP) & = & 0.02
\eeq

\end{document}
