more work added

R/ch2.qmd

@@ -140,13 +140,44 @@ probability table:
 A couple things to note about our table (1) + (3) = .4 and (2) + (4) = .6.
 (1) + (2) + (3) + (4) = 1. 
 
-(1) $P(A \cap B) = P(A|B)P(B)$ we know the likelihood of $L(B|A) = P(A|B)$ and we also 
+(1.) $P(A \cap B) = P(A|B)P(B)$ we know the likelihood of $L(B|A) = P(A|B)$ and we also 
 know the prior so we insert these to get 
 $$ P(A \cap B) = P(A|B)P(B)  = .267 \times .4 = .1068$$
 
-(3) $P(A^c \cap B) = P(A^c|B)P(B)$ in this case we do know the prior $P(B) = .4$, but we 
+(3.) $P(A^c \cap B) = P(A^c|B)P(B)$ in this case we do know the prior $P(B) = .4$, but we 
 don't directly know the value of $P(A^c|B)$, however, we note that $P(A|B) + P(A^c|B) = 1$, 
 therefore we compute $P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733$
 $$ P(A^c \cap B) = P(A^c|B)P(B)  = .733 \times .4 = .2932$$
 
-we now can confirm that $.1068 + .2932 = .4$ 
+we now can confirm that $.1068 + .2932 = .4$ 
+
+Moving on to (2), (4)
+
+(2.) $P(A \cap B^c) = P(A|B^c)P(B^c)$. In this case know the likelihood $L(B^c|A) = P(A|B^c)$ and 
+we know the prior $P(B^c)$ therefore,
+$$P(A \cap B^c) = P(A|B^c)P(B^c) = .022 \times .6 = .0132$$
+
+(4.) $P(A^c \cap B^c) = P(A^c|B^c)P(B^c) = (1 - .022) \times .6 = .5868$
+
+and can confirm that $.0132 + .5868 = .6$
+
+and we can fill the rest of the table:
+
+|      | $B$   | $B^c$  | Total |
+|------|----   |------  |-------|
+|$A$   | .1068 | .0132  | .12   |
+|$A^c$ | .2932 |  .5868 |   .88 |
+|Total | .4    | .6     | 1     |
+
+An important concept we implemented in above is the idea of **total probability**
+
+:::{.callout-tip}
+## total probability
+
+The **total probability** of observing a real article is made up the sum of its
+parts. Namely 
+
+$$P(B^c) = P(A \cap B^c) + P(A^c \cap B^c)$$
+$$=P(A|B^c)P(B^c) + P(A^c|B^c)P(B^c)$$
+$$=.0132 + .5868 = .6$$
+:::