diff --git a/documents/by-course/pstat-120a/course-notes/main.typ b/documents/by-course/pstat-120a/course-notes/main.typ
index 16f8ae0..155529c 100644
--- a/documents/by-course/pstat-120a/course-notes/main.typ
+++ b/documents/by-course/pstat-120a/course-notes/main.typ
@@ -102,8 +102,8 @@ as the notation for $n$ dimensional spaces in $RR$?
 ]
 
 #fact[Generalized DeMorgan's][
-  + $(union_i A_i)' = sect_i A_i '$
-  + $(sect_i A_i)' = union_i A_i '$
+  + $(union.big_i A_i)' = sect.big_i A_i '$
+  + $(sect.big_i A_i)' = union.big_i A_i '$
 ]
 
 == Sizes of infinity
@@ -157,3 +157,257 @@ This gives us the following equivalent statement:
 
   $ N(A) = N(B) <==> exists F : A <-> B $
 ]
+
+= Lecture #datetime(day: 8, month: 1, year: 2025).display()
+
+== Probability
+
+#definition[
+  A *random experiment* is one in which the set of all possible outcomes is known in advance, but one can't predict which outcome will occur on a given trial of the experiment.
+]
+
+#example("Finite sample spaces")[
+  Toss a coin:
+  $ Omega = {H,T} $
+
+  Roll a pair of dice:
+  $ Omega = {1,2,3,4,5,6} times {1,2,3,4,5,6} $
+]
+
+#example("Countably infinite sample spaces")[
+  Shoot a basket until you make one:
+  $ Omega = {M, F M, F F M, F F F M, dots} $
+]
+
+#example("Uncountably infinite sample space")[
+  Waiting time for a bus:
+  $ Omega = {T : t >= 0} $
+]
+
+#fact[
+  Elements of $Omega$ are called sample points.
+]
+
+#definition[
+  Any properly defined subset of $Omega$ is called an *event*.
+]
+
+#example[Dice][
+  Rolling a fair die twice, let $A$ be the event that the combined score of both dice is 10.
+
+  $ A = {(4,6,), (5,5),(6,4)} $
+]
+
+Set theory terms $<-> $ probability terms:
+
+- Superset ($Omega$) $<->$ sample space
+- Element $<->$ outcome / sample point ($omega$)
+- Disjoint sets $<->$ mutually exclusive events
+
+== Classical approach
+
+Classical approach:
+
+$ P(a) = (hash A) / (hash Omega) $
+
+Requires equally likely outcomes and finite sample spaces.
+
+#remark[
+  With an infinite sample space, the probability becomes 0, which is often wrong.
+]
+
+#example("Dice again")[
+  Rolling a fair die twice, let $A$ be the event that the combined score of both dice is 10.
+
+  $
+    A &= {(4,6,), (5,5),(6,4)} \
+    P(A) &= 3 / 36 = 1 / 12
+  $
+]
+
+== Relative frequency approach
+
+$
+  P(A) = (hash "of times" A "occurs in large number of trials") / (hash "of trials")
+$
+
+#example[
+  Flipping a coin to determine the probability of it landing heads.
+]
+
+== Subjective approach
+
+Personal definition of probability.
+
+== Axiomatic approach
+
+Our focus.
+
+#definition[
+  $P(dot)$ is a set function satisfying the 3 axioms
+
+  + $P(A) >= 0, forall A$
+  + $P(Omega) = 1$
+  + If $A_i sect A_j = emptyset, forall i != j$, then
+    $ P(union.big_(i=1)^infinity A_i) = sum_(i=1)^infinity P(A_i) $
+]
+
+#proposition[
+  $ P(emptyset) = 0 $
+]
+
+#proof[
+  By axiom 3,
+
+  $
+    A_1 = emptyset, A_2 = emptyset, A_3 = emptyset \
+    P(emptyset) = sum^infinity_(i=1) P(A_i) = sum^infinity_(i=1) P(emptyset)
+  $
+  Suppose $P(emptyset) != 0$. Then $P >= 0$ by axiom 1 but then $P -> infinity$ in the sum, which implies $Omega > 1$, which is disallowed by axiom 2. So $P(emptyset) = 0$.
+]
+
+#proposition[
+  If $A_1, A_2, ..., A_n$ are disjoint, then
+  $ P(union.big^n_(i=1) A_i) = sum^n_(i= 1) P(A_i) $
+]
+
+#proof[
+  Consider $(A_1, A_2, ..., A_n, emptyset, emptyset, ...)$.
+]
+
+#proposition[Complement][
+  $ P(A') = 1 - P(A) $
+]
+
+#proof[
+  $
+    A' union A &= Omega \
+    A' sect A &= emptyset \
+    P(A' union A) &= P(A') + P(A) &"(by axiom 3)"\
+    = P(Omega) &= 1 &"(by axiom 2)"
+  $
+]
+
+#proposition[
+  $ A subset.eq B => P(A) <= P(B) $
+]
+
+#proof[
+  $ B = A union (A' sect B) $
+
+  but $A$ and ($A' sect B$) are disjoint, so
+
+  $
+    P(B) &= P(A union (A' sect B)) \
+    &= P(A) + P(A' sect B) \
+    &therefore P(B) >= P(A)
+  $
+]
+
+#proposition[
+  $ P(A union B) = P(A) + P(B) - P(A sect B) $
+]
+
+#proof[
+  $
+    A = (A sect B) union (A sect B') \
+    => P(A) = P(A sect B) + P(A sect B') \
+    => P(B) = P(B sect A) + P(B sect A') \
+    P(A) + P(B) = P(A sect B) + P(A sect B) + P(A sect B') + P(A' sect B) \
+    => P(A) + P(B) - P(A sect B) = P(A sect B) + P(A sect B') + P(A' sect B) \
+  $
+]
+
+#example[
+  Select one card from a deck of 52 cards.
+
+  $
+    Omega = {1,2,...,52} \
+    A = "card is a heart" = {H 2, H 3, H 4, ..., H"Ace"} \
+    B = "card is an Ace" = {H"Ace", C"Ace", D"Ace", S"Ace"} \
+    C = "card is black" = {C 2, C 3, ..., C"Ace", S 2, S 3, ..., S"Ace"} \
+    P(A) = 13 / 52,
+    P(B) = 4 / 52,
+    P(C) = 26 / 52 \
+    P(A sect B) = 1 / 52 \
+    P(A sect C) = 0 \
+    P(B sect C) = 2 / 52 \
+    P(A union B) = P(A) + P(B) - P(A sect B) = 16 / 52 \
+    P(B') = 1 - P(B) = 48 / 52 \
+    P(A sect B') = P(A) - P(A sect B) = 13 / 52 - 1 / 52 = 12 / 52 \
+    P((A sect B') union (A' sect B)) = P(A sect B') + P(A' sect B) = 15 / 52 \
+    P(A' sect B') = P(A union B)' = 1 - P(A union B) = 36 / 52
+  $
+]
+
+== Countable sample spaces
+
+#definition[
+  A sample space $Omega$ is said to be *countable* if its finite or countably infinite.
+]
+
+In such a case, one can list the elements of $Omega$.
+
+$ Omega = {omega_1, omega_2, omega_3, ...} $
+with associated probabilities, $p_1, p_2, p_3,...$, where
+$
+  p_i = P(omega_i) >= 0 \
+  1 = P(Omega) = sum P(omega_i)
+$
+
+#example[Fair die, again][
+  All outcomes are equally likely,
+  $ p_1 = p_2 = ... = p_6 = 1 / 6 $
+  Let $A$ be the event that the score is odd = ${1,3,5}$
+  $ P(A) = 3 / 6 $
+]
+
+#example[Loaded die][
+  Consider a die where the probabilities of rolling odd sides is double the probability of rolling an even side.
+  $
+    p_2 = p_4 = p_6, p_1 = p_3 = p_5 = 2p_2 \
+    6p_2 + 3p_2 = 9p_2 = 1 \
+    p_2 = 1 / 9, p_1 = 2 / 9
+  $
+]
+
+#example[Coins][
+  Toss a fair coin until you get the first head.
+  $
+    Omega = {H, T H, T T H, ...} "(countably infinite)" \
+    P(H) = 1 / 2 \
+    P(T T H) = (1 / 2)^3 \
+    P(Omega) = sum_(n=1)^infinity (1 / 2)^n = 1 / (1 - 1 / 2) - 1 = 1
+  $
+]
+
+#example[
+  Birthdays.
+
+  What is the probability two people share the same birthday?
+
+  $
+    Omega = [1,365] times [1,365] \
+    P(A) = 365 / 365^2 = 1 / 365
+  $
+]
+
+== Continuous sample spaces
+
+#definition[
+  A *continuous sample space* contains an interval in $RR$ and is uncountably infinite.
+]
+
+#definition[
+  A probability density function (#smallcaps[pdf]) gives the probability at the point
+  $s$.
+]
+
+Properties of the #smallcaps[pdf]:
+
+- $f(s) >= 0, forall p_i >= 0$
+- $integral_S f(s) dif s = 1, forall p_i >= 0$
+
+#example[
+  Waiting time for bus: $Omega = {s : s >= 0}$.
+]