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From LogicWiki

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 ${\displaystyle A_1,A_2,\dots ,A_n}$ is a collection of independent sentences, then:

${\displaystyle Pr(A_1\vee A_2\vee \dots \vee A_n)=1-Pr(\neg A_1)\cdot Pr(\neg A_2)\cdot \dots \cdot Pr(\neg A_n)}$

Proof of This Trick

To begin, notice that

${\displaystyle A_1\vee A_2\vee \dots \vee A_n}$

is logically equivalent to

${\displaystyle \neg (\neg (A_1\vee A_2\vee \dots \vee A_n))}$

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

${\displaystyle \neg (A_1\vee A_2\vee \dots \vee A_n)}$

is logically equivalent to

${\displaystyle \neg A_1\wedge \neg A_2\wedge \dots \wedge \neg A_n}$

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

${\displaystyle A_1\vee A_2\vee \dots \vee A_n}$

is logically equivalent to

${\displaystyle \neg (\neg A_1\wedge \neg A_2\wedge \dots \wedge \neg A_n)}$

Now we use Rule 3 to show that

${\displaystyle Pr(A_1\vee A_2\vee \dots \vee A_n)=Pr(\neg (\neg A_1\wedge \neg A_2\wedge \dots \wedge \neg A_n))}$

By Rule 5, the right hand side is equal to

${\displaystyle 1-Pr(\neg A_1\wedge \neg A_2\wedge \dots \wedge \neg A_n)}$

Finally, by using Rule 8 several times (and the fact that if ${\displaystyle A_1,A_2,\dots ,A_n}$ are all independent, so are ${\displaystyle \neg A_1,\neg A_2,\dots ,\neg A_n}$; the proof of this is omitted for now), we turn this into

${\displaystyle 1-Pr(\neg A_1)\cdot Pr(\neg A_2)\cdot \dots \cdot Pr(\neg A_n)}$

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.

${\displaystyle A}$ = "Dr. Joseph gets the job at Alaska"

${\displaystyle B}$ = "Dr. Joseph gets the job at Boston College"

${\displaystyle C}$ = "Dr. Joseph gets the job at Columbia"

${\displaystyle D}$ = "Dr. Joseph gets the job at Delaware"

${\displaystyle E}$ = "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 ${\displaystyle Pr(A\vee B\vee C\vee D\vee E)}$. The blurb tells us that ${\displaystyle A,B,C,D,E}$ are all independent, so we can use the trick. We also get that

${\displaystyle Pr(A)=0.25}$

${\displaystyle Pr(B)=0.1}$

${\displaystyle Pr(C)=0.4}$

${\displaystyle Pr(D)=0.1}$

${\displaystyle Pr(E)=0.01}$

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

${\displaystyle Pr(\neg A)=0.75}$

${\displaystyle Pr(\neg B)=0.9}$

${\displaystyle Pr(\neg C)=0.6}$

${\displaystyle Pr(\neg D)=0.9}$

${\displaystyle Pr(\neg E)=0.99}$

The trick tells us that

${\displaystyle Pr(A\vee B\vee C\vee D\vee E)=1-Pr(\neg A)\cdot Pr(\neg B)\cdot Pr(\neg C)\cdot Pr(\neg D)\cdot Pr(\neg E)}$

Which is

${\displaystyle 1-0.75\cdot 0.9\cdot 0.6\cdot 0.9\cdot 0.99=0.639}$

Thus

${\displaystyle Pr(A\vee B\vee C\vee D\vee E)=0.639}$

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

See Also

Independence, Probability Calculus