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Lær at mestre sandsynlighedsregning – Her gennemgår vi Binomialfordelingen, så du kan forstå det på under 2 minutter!

Binomialfordelingen

Binomialfordelingen er en af de emner indenfor sandsynlighedsregningen som mange elever har svært ved. Vi vil derfor i dette indlæg forsøge at præsentere binomialfordelingen på en måde så alle kan være med. Så hvad fortæller Binomialfordelingen os egentlig?

Man kan meget let opsummere hele essensen af binomialfordelingen, den går i sin helhed ud på at man udfører det man kalder et eksperiment n antal gange. Et eksperiment er enten en succes eller en fiasko, der er altså kun to mulige udfald.

Sandsynligheden for at et enkelt eksperiment bliver en succes betegner man for p, og dermed bliver sandsynligheden for fiasko 1-p. I binomialfordelingen har man den stokastiske variabel X, og den angiver det antal gange som et enkelt eksperiment lykkedes.

Lad os prøve at gennemgå et eksempel og på den måde udlede selve binomialfordelingen.

 

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Et eksempel med Binomialfordeling

Eksemplet går ud på at vi kaster en mønt tre gange. Mønten er symmetrisk og derfor er der lige stor chance for at få krone såvel som plat, der er altså 50% chance for at få krone og 50% chance for at få plat.

Vi kaster som sagt mønten tre gange, vi udfører altså eksperimentet tre gange. Vi kalder det for en succes hvis vi får krone og fiasko hvis vi får plat. Vi vil se hvad sandsynligheden er for at få krone 3 gange. Vi skal altså få succes tre gange i streg.

Vi har illustreret de tre kast med et trædiagram som kan ses forneden. Vi kaster mønten en gang. Vores mål er at bestemme sandsynligheden for at få succes tre gange. I første kast er der 50% chance for at få succes, dette har vi betegnet med er (S) i trædiagrammet forneden. Prøv at finde det punkt på træet.

Nu kaster vi mønten endnu en gang og denne gang er chancen igen 50% for at få krone. Vi kan bevæge os videre ud trædiagrammet ved at tage 50% af de allerede fundne 50%.

For at gøre dette skal man benytte multiplikationsprincippet. Vi ganger blot 0,5 med 0,5 og dermed fås 0,25. Det svarer til 25%. Der er altså 25% chance for at få succes to gange i streg. Opnås dette, så ender vi ved punktet (S,S) i trædiagrammet.

 

 

Vi mangler dog nu at kaste mønten en til gang, og igen er der 50% chance for at få succes.

Igen benytter vi multiplikationsprincippet, så denne gang skal vi gange 0,25 med 0,5. Ved at gøre dette fås 0,125. Der er altså 12,5% chance for at få succes tre gange i streg.

Prøv at se de andre udfald i diagrammet og læg mærke til at sandsynligheden er den samme, vær dog opmærksom på at nogle udfald går igen, der er foreksempel tre forskellige måder man kan får to succeser på, man kan få (S,S,F), (S,F,S) eller (F,S,S).

Den samlede sandsynlighed kan vi nu bestemme ved at lægge de tre ”sandsynligheder” sammen. Vi sætter den stokastiske variabel X til at være 2 og vi kan derfor opskrive

Der er altså 37,5% sandsynlighed for at få 2 succeser når man kaster mønten tre gange.

 

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Den måde vi har gjort det på kan måske være lidt træls. Så nu vil vi opskrive et udtryk vi kan bruge til hver en tid, og det udtryk kommer sjovt nok til at være udtrykket for binomialfordelingen som du kender fra din matematikbog.

Vi husker at vi kalder sandsynligheden for succes for p og dermed bliver sandsynligheden for at få en fiasko 1-p. Den stokastiske variabel X kalder vi blot for r. Dermed angiver r det antal gange vi gerne vil have succes. Vi skal derfor opløfte sandsynligheden p i r i vores udtryk.

Dermed starter vi med at have

Vi er dog ikke færdige endnu.

Vi skal nu gange med sandsynligheden for at få fiasko. Fiaskoen kan ske n-r antal gange. Hvor n angiver antallet af eksperimenter vi udfører. Vi skal derfor gange med (1-p)^(n-r). Dette indsætter vi nu i udtrykket foroven

Vi er dog ikke helt i mål endnu.

Vi skal have en ting mere med. Nemlig den at der er flere forskellige måder vores ønskede udfald kan fremkomme. Ligesom i vores kast med mønten, hvor to succeser ud af tre kast kunne fremkomme på tre forskellige måder.

Vi skal derfor kigge lidt på kombinatorikken. Vi skal finde en formel der kan fortælle os hvor mange forskellige udfald vi kan få, når vi har en bestemt r ud af et bestemt antal n.

Den formel har formen

Vi tilføjer nu ovenstående formel til vores udtryk for binomialfordelingen

Vi er nu nået i mål og har nu udledt binomialfordelingen.

 

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Men inden vi afslutter, så lad os lige teste formlen og med vores mønt-eksempel og se om vi får 0,375 når vi sætter r = 2 og har i alt 3 kast. Vi sætter altså r = 2, n = 3. Sandsynligheden for succes er 0,5, vi sætter derfor p = 0,5.

Dette indsætter vi nu i udtrykket for binomialfordelingen

Vi udregner nu ovenstående udtryk og dermed fås

Og dermed bliver sandsynligheden 37,5% for at få to successer når man kaster en mønt tre gange.

Vil du blive bedre til emner i matematikken, så har jeg samlet endnu flere artikler til dig herunder 🙂

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