LGT

1

Consider joint probability table:

	$p$	$\neg p$

	$q$	$\neg q$	$q$	$\neg q$
$r$	0.008	0.182	0.002	0.328
$\neg r$	0.169	0.044	0.231	0.036

Calculate $P(\neg q,r|\neg p)\cdot P(p,q,r)$.

$$\begin{array}{rl}P(A|B)& =\frac{P(A,B)}{P(B)}\\ & \\ P(\neg q,r|\neg p)& =\frac{P(\neg q,r,\neg p)}{P(\neg p)}\\ P(\neg q,r,\neg p)& =0.328\\ P(\neg p)& =0.002+0.328+0.231+0.036\end{array}$$

this will be on keats later (???)

2

Bayes' Theorem

Bayes’ Theorem states that $$ p(F | E) = \frac{p(E|F)p(F)}{p(E|F)p(F) + p(E|\bar F)p(\bar F)} $$

Short form: $P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$

Link to original

Jar 1 contains 30 vanilla cookies, 10 chocolate cookies Jar 2 contains 20 vanilla cookies, 20 chocolate cookies

After picking at random, you’re told you’ve selected a vanilla cookie What is probability the cookie came from jar 1?

Let $J$ be that we picked from the first jar Let $V$ be that we picked a vanilla cookie

$$\begin{array}{rl}P(J|V)& =\frac{P(V|J)\cdot P(J)}{P(V)}\\ & =\frac{0.75\cdot 0.5}{P(V|J)\cdot P(J)+P(V|\bar{J})\cdot P(\bar{J})}\\ & =\frac{}{}\\ & =0.6\end{array}$$

3

Prove that the following two statements are equivalent

$$\begin{array}{rl}1.P(A|B,C)& =P(A|C)\\ 2.P(A,B|C)& =P(A|C)\cdot P(B|C)\end{array}$$

$$\begin{array}{rl}P(A|B,C)& =\frac{P(A,B,C)}{P(B,C)}\\ & =\frac{P(A,B|C)\cdot \cancel{P(C)}}{P(B|C)\cdot \cancel{P(C)}}\\ & =\frac{P(A,B|C)}{P(B|C)}\\ & =\frac{P(A|C)\cdot \cancel{P(B|C)}}{\cancel{P(B|C)}}\\ & =P(A|C)\end{array}$$

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