Rules and Properties of Probability

Complement Rule

For any event $A$ , $P (A) = 1 - P (A^{'})$

Proof
By definition of $A^{'}$ , $A \cup A^{'} = S$ so then since $A, A^{'}$ are disjoint we have:

1 = P (S) = P (A \cup A^{'}) = P (A) + P (A^{'})

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A Series System

Consider a series-connected system of 5 components. Say $A$ is the event that the system fails. For $A$ to happen, just one component must fail. We have:

A = {F F F F F, F F F F S, . . ., F S S S S} = (A^{'})^{'} = {S S S S S}^{'}

Clearly since $P (A^{'}) = \frac{1}{2^{5}} = \frac{1}{32}$ then $P (A) = \frac{31}{32}$ .

P (A) \leq 1

For any event $A$ , $P (A) \leq 1$ .

Proof

Using the complement rule: $1 = P (A) + P (A^{'}) \geq P (A)$ since $P (A^{'}) \geq 0$ by Axiom 2.
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Addition Rule

For any events $A, B$ :

P(A \cup B) = P(A) + P(B) - P(A \cap B) $${#70bda0}

Proof
Note that $A \cup B = A \cup (B \cap A^{'})$ , which is used without too much proof.
Pasted image 20240924092909.png

Because $A$ and $B \cap A^{'}$ are disjoint, then $P (A \cup B) = P (A) + P (B \cap A^{'})$ using the complement rule. But $B = (B \cap A) \cup (B \cap A^{'})$ , so then since $B \cap A$ and $B \cap A^{'}$ are disjoint then $P (B) = P (B \cap A) + P (B \cap A^{'})$ . Combining these two:

P (A \cup B) = P (A) + P (B \cap A^{'}) = P (A) + (P (B) - P (B \cap A))

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These properties generally extend for 3 or more sets as well. This patter of additions and subtractions is called the inclusion-exclusion principle:

\begin{array}{r} P (A \cup B \cup C) = \\ P (A) + P (B) + P (C) - P (A \cap B) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C) \end{array}

Example

Residential Probabilities

If a suburb has 60% of residents with internet service, 80% with television service, and 50% with both, what is the probability that a randomly selected household has at least one of these services? Probability of exactly one service?

Proof
Here let $I$ be the internet service event, and $T$ for television. Here $P (I) = 0.6$ and $P (T) = 0.8$ . Consequently $P (I \cap T) = 0.5$ . Using the addition rule:

\begin{aligned} P (at least one service) & = P (I \cup T) \\ = P (I) + P (T) - P (I \cap T) \\ = 0.6 + 0.8 - 0.5 \\ = 0.9 \end{aligned}

Similarly for exactly one service:

\begin{aligned} P (exactly one service) & = P (I \cap T^{'}) + P (I^{'} \cap T) \end{aligned}

\begin{aligned} P (I \cap T^{'}) + P (I \cap T) + P (I^{'} \cap T) + P (no service) & = 1 \\ P (exactly one) + P (I \cap T) + P (I^{'} \cap T^{'}) & = 1 \\ P (exactly one) + 0.5 + 0.1 & = 1 \\ \Rightarrow P (exactly one) & = 0.4 \end{aligned}

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References

[[Matthew A. Carlton, Jay L. Devore - Probability with STEM Applications-Wiley (2020).pdf#page=46]]