Stat 302. Ruben Zamar Inclusion-Exclusion Formula. Ruben Zamar () Module 6 Inclusion-Exclusion Formula 1 / 8
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1 Stat 302 Rube Zamar Iclusio-Exclusio Formula Rube Zamar () Module 6 Iclusio-Exclusio Formula 1 / 8
2 Iclusio-Exclusio Idetity (3 Evets) We have P (E F G ) = P (E ) + P (F ) + P (G ) P (EF ) P (EG ) P (FG ) +P (EFG ). Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 2 / 8
3 Iclusio-Exclusio Idetity (3 Evets) We have P (E F G ) = P (E ) + P (F ) + P (G ) P (EF ) P (EG ) P (FG ) +P (EFG ). Proof. Notice that We will use the formula E F G = (E F ) G P (A B) = P (A) + P (B) P (AB) with A = E F, B = G Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 2 / 8
4 Proof (Cotiued) Now P (E F G ) = P (E F ) + P (G ) P ((E F ) G ) = P (E ) + P (F ) P (EF ) + P (G ) P ((E F ) G ). Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 3 / 8
5 Proof (Cotiued) Now P (E F G ) = P (E F ) + P (G ) P ((E F ) G ) = P (E ) + P (F ) P (EF ) + P (G ) P ((E F ) G ). Moreover (E F ) G = (EG ) (FG ) so P ((E F ) G ) = P (EG ) + P (FG ) P ((EG ) (FG )) ad the result follows. = P (EG ) + P (FG ) P (EFG ) ube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 3 / 8
6 Geeral Iclusio-Exclusio Idetity We have P (E 1 E 2 E ) = = i=1 P (E i ) P (E i1 E i2 ) + i 1 <i 2 + ( 1) r +1 i 1 <i 2 < <i r P (E i1 E ir ) + + ( 1) +1 P (E 1 E ) ( 1) r +1 P (E i1 E ir ) r =1 i 1 <i 2 < <i r Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 4 / 8
7 Geeral Iclusio-Exclusio Idetity We have P (E 1 E 2 E ) = = i=1 P (E i ) P (E i1 E i2 ) + i 1 <i 2 + ( 1) r +1 i 1 <i 2 < <i r P (E i1 E ir ) + + ( 1) +1 P (E 1 E ) ( 1) r +1 P (E i1 E ir ) r =1 i 1 <i 2 < <i r This ca be prove by iductio. Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 4 / 8
8 Example: Matchig problem You have letters ad evelopes ad radomly stuff the letters i the evelopes. Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 5 / 8
9 Example: Matchig problem You have letters ad evelopes ad radomly stuff the letters i the evelopes. What is the probability that at least oe letter will match its iteded evelope? Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 5 / 8
10 Example: Matchig problem You have letters ad evelopes ad radomly stuff the letters i the evelopes. What is the probability that at least oe letter will match its iteded evelope? The sample space is the space of permutatios of {1, 2,..., } ad thus has! outcomes. Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 5 / 8
11 Example: Matchig problem You have letters ad evelopes ad radomly stuff the letters i the evelopes. What is the probability that at least oe letter will match its iteded evelope? The sample space is the space of permutatios of {1, 2,..., } ad thus has! outcomes. Let E i = { letter i matches its iteded evelope } Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 5 / 8
12 Example: Matchig problem You have letters ad evelopes ad radomly stuff the letters i the evelopes. What is the probability that at least oe letter will match its iteded evelope? The sample space is the space of permutatios of {1, 2,..., } ad thus has! outcomes. Let E i = { letter i matches its iteded evelope }. We are iterested i P (E 1 E 2 E ). Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 5 / 8
13 Matchig problem (Cotiued) Cosider evets of the form E i1 E ir Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 6 / 8
14 Matchig problem (Cotiued) Cosider evets of the form E i1 E ir That is, the r letters i 1,..., i r match their iteded evelopes. Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 6 / 8
15 Matchig problem (Cotiued) Cosider evets of the form E i1 E ir That is, the r letters i 1,..., i r match their iteded evelopes. There are ( r) ( r 1) 1 = ( r)! such outcomes correspodig to the umber of ways the remaiig r evelopes ca be matched. Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 6 / 8
16 Matchig problem (Cotiued) Cosider evets of the form E i1 E ir That is, the r letters i 1,..., i r match their iteded evelopes. There are ( r) ( r 1) 1 = ( r)! such outcomes correspodig to the umber of ways the remaiig r evelopes ca be matched. Assumig all outcomes are equaly likely P (E i1 E ir ) = ( r) ( r 1) 1! = ( r)!! Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 6 / 8
17 Matchig problem (cot.) We wat to compute P (E 1 E 2 E ) = ( 1) r +1 P (E i1 E ir ) r =1 i 1 <i 2 < <i r Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 7 / 8
18 Matchig problem (cot.) We wat to compute P (E 1 E 2 E ) = We have ( r ( 1) r +1 P (E i1 E ir ) r =1 i 1 <i 2 < <i r ) terms of the form P (E i1 E ir ). Moreover, ( r ) P (E i1 E ir ) = 1 }{{} r!. ( r )! =! Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 7 / 8
19 Matchig problem (cot.) It follows that P (E 1 E 2 E ) = ( 1) r +1 1 r =1 r! = 1 1 2! + 1 3! + ( 1)+1! 1 e 1 as Rube Zamar rube@stat.ubc.ca () Module 6 Iclusio-Exclusio Formula 8 / 8
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