MATHHX B

MATHHX B

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{\morecmidrules }{}\) \(\newcommand {\specialrule }[3]{\hline }\) \(\newcommand {\addlinespace }[1][]{}\) \(\def \LWRsiunitxrangephrase { \protect \mbox {to (numerical range)} }\) \(\def \LWRsiunitxdecimal {.}\)

14.3 Beviser til binomialfordeling

  • Sætning 14.1.1
    For en binomialfordeling \(X\sim b(n,p)\) gælder:

    \[P(X=r)=K(n,r)\cdot p^r \cdot (1-p)^{n-r}\]

  • Bevis 
    Vi vil vise sætningen gennem et eksempel. Det er normalt ikke noget man accepterer i et bevis, da man jo så i princippet ikke kan være sikker på, at sætningen gælder i alle tilfælde. Derfor skal vi hele tiden have i tankerne, at det vi gør også skal virke i andre tilfælde (for alle værdier af \(n\) og \(r\)).

    Vi kigger på tilfældet hvor \(n=5\) og \(r=2\). Vi skal altså have 2 succeser i 5 forsøg. Det kan vi få på flere måder. Hvis vi kalder success for \(s\) og fiasko for \(f\) kunne det f.eks. se sådan ud.

    .
    Forsøg nr. 1 2 3 4 5
    Succes/fiasko s s f f f

    Det kunne også se sådan ud:

    .
    Forsøg nr. 1 2 3 4 5
    Succes/fiasko f s s f f

    Eller hvad med sådan?

    .
    Forsøg nr. 1 2 3 4 5
    Succes/fiasko s f f s f

    Der er flere muligheder...

    Vi vil nu se på sandsynlighederne for de 3 nævnte kombinationer af succes og fiasko. Vi ved at sandsynligheden for succes er \(p\) og derfor må sandsynligheden for fiasko være \(1-p\) (fiasko er komplementær til succes). Vi ved også at forsøgene er uafhængige (vi har forudsat at sandsynligheden er fast hver gang), hvilket betyder at vi kan gange sandsynlighederne sammen for at få den samlede sandsynlighed. Derfor må sandsynligheden for det første eksempel være givet ved

    \[p\cdot p\cdot (1-p)\cdot (1-p)\cdot (1-p)=p^2\cdot (1-p)^3.\]

    Det næste eksempel må have sandsynligheden

    \[(1-p)\cdot p\cdot p\cdot (1-p)\cdot (1-p)=p^2\cdot (1-p)^3.\]

    Det sidste eksempel må have sandsynligheden

    \[p\cdot (1-p)\cdot (1-p)\cdot p\cdot (1-p)=p^2\cdot (1-p)^3.\]

    Ved at tænke os lidt om kan vi godt se, at alle de forskellige mulige kombinationer af 2 successer og 3 fiaskoer må have samme sandsynlighed nemlig \(p^2\cdot (1-p)^3\).

    Vi mangler nu bare at finde ud af hvor mange måder vi kan få de 2 succeser i de 5 forsøg. Som det ses af tabellerne, er det det samme som at spørge, hvor mange måder man kan placere 2 krydser på 5 pladser og det lærte vi i kombinatorik. Det var jo tallet \(K(5,2)\). Altså kan vi få 2 succeser i 5 forsøg på \(K(5,2)\) forskellige måder og hver af disse kombinationer har sandsynligheden \(p^2\cdot (1-p)^3\).

    Vi kan nu finde dem samlede sandsynlighed \(P(X=2)\) ved at gange antallet af kombinationer \(K(5,2)\) med sandsynligheden \(p^2\cdot (1-p)^3\) for hver af kombinationerne (da sandsynligheden for en hændelse er lig med summen af sandsynlighederne for de enkelte udfald). Vi får så

    \[P(X=2)=K(5,2)\cdot p^2 \cdot (1-p)^3.\]

    Vi husker nu hvor tallene 2, 3 og 5 kom fra. Tallet 2 kom fra at vi ville have to succeser, så \(r=2\). Tallet 5 kom fra at vi havde 5 forsøg, dvs. \(n=5\). Tallet 3 kom fra at vi havde 3 fiaskoer og det fandt vi ud af ved at sige \(5-2=n-r\). Derfor må vi have:

    \[P(X=r)=K(n,r)\cdot p^r \cdot (1-p)^{n-r}.\]