auto-update(nvim): 2025-01-08 15:43:40
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@ -102,8 +102,8 @@ as the notation for $n$ dimensional spaces in $RR$?
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]
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#fact[Generalized DeMorgan's][
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+ $(union_i A_i)' = sect_i A_i '$
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+ $(sect_i A_i)' = union_i A_i '$
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+ $(union.big_i A_i)' = sect.big_i A_i '$
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+ $(sect.big_i A_i)' = union.big_i A_i '$
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]
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== Sizes of infinity
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@ -157,3 +157,257 @@ This gives us the following equivalent statement:
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$ N(A) = N(B) <==> exists F : A <-> B $
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]
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= Lecture #datetime(day: 8, month: 1, year: 2025).display()
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== Probability
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#definition[
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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.
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]
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#example("Finite sample spaces")[
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Toss a coin:
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$ Omega = {H,T} $
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Roll a pair of dice:
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$ Omega = {1,2,3,4,5,6} times {1,2,3,4,5,6} $
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]
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#example("Countably infinite sample spaces")[
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Shoot a basket until you make one:
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$ Omega = {M, F M, F F M, F F F M, dots} $
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]
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#example("Uncountably infinite sample space")[
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Waiting time for a bus:
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$ Omega = {T : t >= 0} $
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]
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#fact[
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Elements of $Omega$ are called sample points.
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]
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#definition[
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Any properly defined subset of $Omega$ is called an *event*.
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]
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#example[Dice][
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Rolling a fair die twice, let $A$ be the event that the combined score of both dice is 10.
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$ A = {(4,6,), (5,5),(6,4)} $
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]
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Set theory terms $<-> $ probability terms:
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- Superset ($Omega$) $<->$ sample space
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- Element $<->$ outcome / sample point ($omega$)
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- Disjoint sets $<->$ mutually exclusive events
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== Classical approach
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Classical approach:
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$ P(a) = (hash A) / (hash Omega) $
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Requires equally likely outcomes and finite sample spaces.
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#remark[
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With an infinite sample space, the probability becomes 0, which is often wrong.
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]
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#example("Dice again")[
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Rolling a fair die twice, let $A$ be the event that the combined score of both dice is 10.
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$
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A &= {(4,6,), (5,5),(6,4)} \
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P(A) &= 3 / 36 = 1 / 12
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$
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]
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== Relative frequency approach
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$
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P(A) = (hash "of times" A "occurs in large number of trials") / (hash "of trials")
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$
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#example[
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Flipping a coin to determine the probability of it landing heads.
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]
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== Subjective approach
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Personal definition of probability.
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== Axiomatic approach
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Our focus.
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#definition[
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$P(dot)$ is a set function satisfying the 3 axioms
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+ $P(A) >= 0, forall A$
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+ $P(Omega) = 1$
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+ If $A_i sect A_j = emptyset, forall i != j$, then
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$ P(union.big_(i=1)^infinity A_i) = sum_(i=1)^infinity P(A_i) $
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]
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#proposition[
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$ P(emptyset) = 0 $
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]
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#proof[
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By axiom 3,
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$
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A_1 = emptyset, A_2 = emptyset, A_3 = emptyset \
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P(emptyset) = sum^infinity_(i=1) P(A_i) = sum^infinity_(i=1) P(emptyset)
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$
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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 $P(emptyset) = 0$.
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]
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#proposition[
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If $A_1, A_2, ..., A_n$ are disjoint, then
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$ P(union.big^n_(i=1) A_i) = sum^n_(i= 1) P(A_i) $
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]
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#proof[
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Consider $(A_1, A_2, ..., A_n, emptyset, emptyset, ...)$.
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]
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#proposition[Complement][
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$ P(A') = 1 - P(A) $
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]
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#proof[
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$
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A' union A &= Omega \
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A' sect A &= emptyset \
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P(A' union A) &= P(A') + P(A) &"(by axiom 3)"\
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= P(Omega) &= 1 &"(by axiom 2)"
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$
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]
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#proposition[
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$ A subset.eq B => P(A) <= P(B) $
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]
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#proof[
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$ B = A union (A' sect B) $
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but $A$ and ($A' sect B$) are disjoint, so
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$
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P(B) &= P(A union (A' sect B)) \
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&= P(A) + P(A' sect B) \
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&therefore P(B) >= P(A)
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$
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]
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#proposition[
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$ P(A union B) = P(A) + P(B) - P(A sect B) $
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]
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#proof[
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$
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A = (A sect B) union (A sect B') \
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=> P(A) = P(A sect B) + P(A sect B') \
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=> P(B) = P(B sect A) + P(B sect A') \
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P(A) + P(B) = P(A sect B) + P(A sect B) + P(A sect B') + P(A' sect B) \
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=> P(A) + P(B) - P(A sect B) = P(A sect B) + P(A sect B') + P(A' sect B) \
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$
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]
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#example[
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Select one card from a deck of 52 cards.
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$
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Omega = {1,2,...,52} \
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A = "card is a heart" = {H 2, H 3, H 4, ..., H"Ace"} \
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B = "card is an Ace" = {H"Ace", C"Ace", D"Ace", S"Ace"} \
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C = "card is black" = {C 2, C 3, ..., C"Ace", S 2, S 3, ..., S"Ace"} \
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P(A) = 13 / 52,
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P(B) = 4 / 52,
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P(C) = 26 / 52 \
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P(A sect B) = 1 / 52 \
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P(A sect C) = 0 \
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P(B sect C) = 2 / 52 \
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P(A union B) = P(A) + P(B) - P(A sect B) = 16 / 52 \
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P(B') = 1 - P(B) = 48 / 52 \
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P(A sect B') = P(A) - P(A sect B) = 13 / 52 - 1 / 52 = 12 / 52 \
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P((A sect B') union (A' sect B)) = P(A sect B') + P(A' sect B) = 15 / 52 \
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P(A' sect B') = P(A union B)' = 1 - P(A union B) = 36 / 52
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$
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]
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== Countable sample spaces
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#definition[
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A sample space $Omega$ is said to be *countable* if its finite or countably infinite.
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]
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In such a case, one can list the elements of $Omega$.
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$ Omega = {omega_1, omega_2, omega_3, ...} $
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with associated probabilities, $p_1, p_2, p_3,...$, where
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$
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p_i = P(omega_i) >= 0 \
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1 = P(Omega) = sum P(omega_i)
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$
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#example[Fair die, again][
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All outcomes are equally likely,
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$ p_1 = p_2 = ... = p_6 = 1 / 6 $
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Let $A$ be the event that the score is odd = ${1,3,5}$
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$ P(A) = 3 / 6 $
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]
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#example[Loaded die][
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Consider a die where the probabilities of rolling odd sides is double the probability of rolling an even side.
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$
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p_2 = p_4 = p_6, p_1 = p_3 = p_5 = 2p_2 \
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6p_2 + 3p_2 = 9p_2 = 1 \
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p_2 = 1 / 9, p_1 = 2 / 9
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$
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]
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#example[Coins][
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Toss a fair coin until you get the first head.
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$
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Omega = {H, T H, T T H, ...} "(countably infinite)" \
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P(H) = 1 / 2 \
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P(T T H) = (1 / 2)^3 \
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P(Omega) = sum_(n=1)^infinity (1 / 2)^n = 1 / (1 - 1 / 2) - 1 = 1
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$
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]
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#example[
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Birthdays.
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What is the probability two people share the same birthday?
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$
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Omega = [1,365] times [1,365] \
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P(A) = 365 / 365^2 = 1 / 365
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$
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]
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== Continuous sample spaces
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#definition[
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A *continuous sample space* contains an interval in $RR$ and is uncountably infinite.
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]
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#definition[
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A probability density function (#smallcaps[pdf]) gives the probability at the point
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$s$.
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]
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Properties of the #smallcaps[pdf]:
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- $f(s) >= 0, forall p_i >= 0$
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- $integral_S f(s) dif s = 1, forall p_i >= 0$
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#example[
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Waiting time for bus: $Omega = {s : s >= 0}$.
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]
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