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# Disjunction of independent sentences

There is a handy trick for calculating the probability of a disjunction of independent sentences. This is useful because many common problems involve the calculation of the probability of a disjunction of independent sentences, as can be seen in the examples below.

If ${A}_{1},{A}_{2},\dots ,{A}_{n}$ is a collection of independent sentences, then:

 $P\phantom{\rule{0}{0ex}}r\left({A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}\right)=1-P\phantom{\rule{0}{0ex}}r\left(¬{A}_{1}\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬{A}_{2}\right)\cdot \dots \cdot P\phantom{\rule{0}{0ex}}r\left(¬{A}_{n}\right)$

## Proof of This Trick

To begin, notice that

${A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}$

$¬\left(¬\left({A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}\right)\right)$

This is a simple fact of negation. We also have that

$¬\left({A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}\right)$

$¬{A}_{1}\wedge ¬{A}_{2}\wedge \dots \wedge ¬{A}_{n}$

This follows from repeated applications of De Morgan's Law. So we now have that

${A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}$

$¬\left(¬{A}_{1}\wedge ¬{A}_{2}\wedge \dots \wedge ¬{A}_{n}\right)$

Now we use Rule 3 to show that

$P\phantom{\rule{0}{0ex}}r\left({A}_{1}\vee {A}_{2}\vee \dots \vee {A}_{n}\right)=P\phantom{\rule{0}{0ex}}r\left(¬\left(¬{A}_{1}\wedge ¬{A}_{2}\wedge \dots \wedge ¬{A}_{n}\right)\right)$

By Rule 5, the right hand side is equal to

$1-P\phantom{\rule{0}{0ex}}r\left(¬{A}_{1}\wedge ¬{A}_{2}\wedge \dots \wedge ¬{A}_{n}\right)$

Finally, by using Rule 8 several times (and the fact that if ${A}_{1},{A}_{2},\dots ,{A}_{n}$ are all independent, so are $¬{A}_{1},¬{A}_{2},\dots ,¬{A}_{n}$; the proof of this is omitted for now), we turn this into

$1-P\phantom{\rule{0}{0ex}}r\left(¬{A}_{1}\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬{A}_{2}\right)\cdot \dots \cdot P\phantom{\rule{0}{0ex}}r\left(¬{A}_{n}\right)$

This completes the proof of the trick.

## Examples

Dr. Joseph is looking for a job. He has applied to five schools: Alaska, Boston College, Columbia, Delaware, and Eastern Michigan. There are four applicants for the Alaska position, each with an equal chance of getting the job. At Boston College, he only has a 10% chance, while at Columbia he is favored with a 40% chance. Delaware has ten applicants, all of which have an equal chance. Finally, he has a slim one percent chance at the Eastern Michigan job. Assuming that the hiring decisions are made independently of each other, what is the probability that Dr. Joseph gets at least one job offer?

First off, some abbreviations.

$A$ = "Dr. Joseph gets the job at Alaska"

$B$ = "Dr. Joseph gets the job at Boston College"

$C$ = "Dr. Joseph gets the job at Columbia"

$D$ = "Dr. Joseph gets the job at Delaware"

$E$ = "Dr. Joseph gets the job at Eastern Michigan"

The question asks for the probability that Dr. Joseph gets at least one job offer. "At least one" means disjunction, so what we're looking for is $P\phantom{\rule{0}{0ex}}r\left(A\vee B\vee C\vee D\vee E\right)$. The blurb tells us that $A,B,C,D,E$ are all independent, so we can use the trick. We also get that

$P\phantom{\rule{0}{0ex}}r\left(A\right)=0.25$

$P\phantom{\rule{0}{0ex}}r\left(B\right)=0.1$

$P\phantom{\rule{0}{0ex}}r\left(C\right)=0.4$

$P\phantom{\rule{0}{0ex}}r\left(D\right)=0.1$

$P\phantom{\rule{0}{0ex}}r\left(E\right)=0.01$

To use the trick, we need the probability of the negations of these sentences. By using Rule 5, we get:

$P\phantom{\rule{0}{0ex}}r\left(¬A\right)=0.75$

$P\phantom{\rule{0}{0ex}}r\left(¬B\right)=0.9$

$P\phantom{\rule{0}{0ex}}r\left(¬C\right)=0.6$

$P\phantom{\rule{0}{0ex}}r\left(¬D\right)=0.9$

$P\phantom{\rule{0}{0ex}}r\left(¬E\right)=0.99$

The trick tells us that

$P\phantom{\rule{0}{0ex}}r\left(A\vee B\vee C\vee D\vee E\right)=1-P\phantom{\rule{0}{0ex}}r\left(¬A\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬B\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬C\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬D\right)\cdot P\phantom{\rule{0}{0ex}}r\left(¬E\right)$

Which is

$1-0.75\cdot 0.9\cdot 0.6\cdot 0.9\cdot 0.99=0.639$

Thus

$P\phantom{\rule{0}{0ex}}r\left(A\vee B\vee C\vee D\vee E\right)=0.639$

So there is a 63.9% chance that Dr. Joseph will get at least one job offer.