Independence and Probabilities of Independent Events

A catastrophe bond pays out if both a Florida hurricane (probability 0.08) and a California earthquake (0.03) strike the same year. If independent, the joint probability is $0.08\times 0.03=0.0024$. Mistakenly assume dependence and you misprice by an order of magnitude.

Two events $A$ and $B$ are independent if and only if:

$$P(A\cap B)=P(A)\cdot P(B)$$

Equivalently: $P(A\mid B)=P(A)$. Knowing $B$ occurred gives you zero information about $A$.

HIGH-FREQUENCY: The exam tests independence in two directions: (1) given that events are independent, compute joint or conditional probabilities; (2) given numerical values, check whether $P(A\cap B)=P(A)\cdot P(B)$.

This is the most confused pair of concepts in introductory probability.

If $P(A)>0$ and $P(B)>0$, mutually exclusive events are never independent. Knowing one occurred eliminates the other, that is maximum dependence, not independence.

Common mistakes

• Confusing independence with mutual exclusivity. If $P(A)=0.3$, $P(B)=0.5$, and the events are independent, then $P(A\cap B)=0.15$. Writing 0 (mutually exclusive) is wrong. Trap: 0 instead of 0.15.
• Adding instead of multiplying for joint independent events. $P(F\cap L\cap T)=0.04\times 0.02\times 0.06=0.000048$, not $0.04+0.02+0.06=0.12$. Trap: 0.12.
• Using $1-\prod p_i$ for "at least one." The correct formula is $1-\prod (1-p_i)$. Subtracting the product of the probabilities themselves is wrong. Trap: $1-0.96\times 0.98\times 0.94\ne$ the right answer for at least one (it is for "not all three").

Bottom line

• Independence definition: $P(A\cap B)=P(A)\cdot P(B)$, equivalently $P(A\mid B)=P(A)$; the product and conditional tests are algebraically identical
• Mutually exclusive events with positive probabilities are NEVER independent (a zero joint probability cannot equal a positive product)
• Complement swap: if $A,B$ independent, so are $A,B^c$ and $A^c,B^c$ (all four combinations)
• Mutual independence of $n$ events needs $2^n-n-1$ conditions; pairwise independence does NOT let you factor the triple intersection

Exam shortcut

When a problem says "independent," immediately write $P(A\cap B)=P(A)P(B)$. When asked "are these independent?", compute the product and compare to the given intersection. For reliability, draw the series-parallel diagram first. "Independent = Irrelevant." Knowing B happened is irrelevant to A. Mutual exclusivity is the opposite, knowing B eliminates A. "Series Shrinks, Parallel Protects." Series reliability is always less than the weakest component. Parallel is always greater than the strongest.

The full lesson (about 4,109 words, 27 min read) adds 6 worked examples, all 8 common mistakes, a self-check, free in the app.

Learning objectives

• 1c

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