fixed formatting for 1I3

_problems/unit-01/I-law-of-total-probability-and-Bayes-rule/3.md

@@ -2,7 +2,8 @@
 index: 3
 level: 1
 statement: |
-    A test for coronaviruses has the following properties:    Given that a patient has the coronavirus, the conditional probability of a positive test is .95.
+    A test for coronaviruses has the following properties:
+    Given that a patient has the coronavirus, the conditional probability of a positive test is .95.
     Given that a patient does not have the coronavirus, the conditional probability of a negative test is .95
     You believe that in the population each patient has a .10 probability of having coronavirus.
     What is the probability that a person has coronavirus, given they present a positive result? 
@@ -11,13 +12,13 @@ Let $T$ be the event that a **t**est comes up positive.  Let $C$ be the event th
 
 The problem tells us that
 
-- $$P(C) = 0.1$$
-- $$P(T|C) = .95$$
-- $$P(T^C|C^C) = .95$$
+- $P(C) = 0.1$
+- $P(T|C) = .95$
+- $P(T^C|C^C) = .95$
 
-First, by the complement rule,  $$P(T|C^C) = 1 -  P(T^C|C^C) = 1 - .95 = .05$$. Also by the complement rule, $$P(C^C) = 1-P(C) = 1 - 0.1 = 0.9$$
+First, by the complement rule,  $P(T|C^C) = 1 -  P(T^C|C^C) = 1 - .95 = .05$. Also by the complement rule, $P(C^C) = 1-P(C) = 1 - 0.1 = 0.9$
 
-To apply Bayes' Theorem, we first need the probability of a positive test. We notice that $$\{C, C^C\}$$ is a partition of the sample space. Writing the law of total probability,
+To apply Bayes' Theorem, we first need the probability of a positive test. We notice that $\{C, C^C\}$ is a partition of the sample space. Writing the law of total probability,
 
 $$P(T) = P(C)P(T|C) + P(C^C)P(T|C^C) = (0.1)(0.95) + (0.9)(0.05) = 0.14$$