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@@ -9,24 +9,28 @@ A **sample space** $S$ is the set of all possible outcomes, and an **event** is
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## Probability measure axioms
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A probability measure $P$ assigns numbers to events such that $0\le P(E)\le 1$, $P(\emptyset)=0$, $P(S)=1$, and $P(E_1\cup E_2\cup\cdots)=P(E_1)+P(E_2)+\cdots$ for disjoint events. Probabilities can be interpreted as areas in a Venn diagram.
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A probability measure $P$ assigns numbers to events such that $0\le P(E)\le 1$, $P(\emptyset)=0$, $P(S)=1$, and $P(E_1\cup E_2\cup\cdots)=P(E_1)+P(E_2)+\cdots$ for disjoint events. Probabilities can be interpreted as areas in a Venn diagram.
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### Identities and the inclusion–exclusion principle
- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and more generally for $n$ events the inclusion–exclusion formula sums intersections of increasing size with alternating signs. The union bound states $P(E_1\cup\cdots\cup E_n)\le P(E_1)+\cdots+P(E_n)$.
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- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and more generally for $n$ events the inclusion–exclusion formula sums intersections of increasing size with alternating signs. The union bound states $P(E_1\cup\cdots\cup E_n)\le P(E_1)+\cdots+P(E_n)$.
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## Uniform probability on finite spaces
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If $S$ has finitely many equally likely outcomes, then $P(E)=\frac{\#E}{\#S}$.
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> Example: two dice have $36$ equally likely outcomes; the event “sum = 7” contains six outcomes, so the probability is $6/36=1/6$. Drawing 7 cards from a 52‑card deck, the probability of getting a four‑of‑a‑kind is
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If $S$ has finitely many equally likely outcomes, then
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$$
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P(E)=\frac{13\cdot {48\choose 3}}{{52\choose 7}}.
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P(E)=\frac{\text{Number of elements in } E}{\text{Number of elements in } S}.
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$$
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> Example: two dice have $36$ equally likely outcomes; the event “sum = 7” contains six outcomes, so the probability is $6/36=1/6$. Drawing 7 cards from a 52‑card deck, the probability of getting a four‑of‑a‑kind is
In the birthday problem with 23 people, the probability that at least two share a birthday is $1-\frac{365\cdot364\cdots(365-22)}{365^{23}}$, which exceeds $\tfrac{1}{2}$.
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