A couple things to note about our table (1) + (2) = .4 and (2) + (4) = .6. (1) + (2) + (3) + (4) = 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 \[ P(A \cap B) = P(A|B)P(B) = .267 \times .4 = .1068\]
\(P(A^c \cap B) = P(A^c|B)P(B)\) 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 \(P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733\) \[ P(A^c \cap B) = P(A^c|B)P(B) = .733 \times .4 = .2932\]
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 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 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 \(.1068 + .2932 = .4\)
+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
+ diff --git a/R/ch2.qmd b/R/ch2.qmd index 17fd6ac..ef93a80 100644 --- a/R/ch2.qmd +++ b/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 $.1068 + .2932 = .4$ \ No newline at end of file +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 + +$$P(B^c) = P(A \cap B^c) + P(A^c \cap B^c)$$ +$$=P(A|B^c)P(B^c) + P(A^c|B^c)P(B^c)$$ +$$=.0132 + .5868 = .6$$ +::: \ No newline at end of file