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Waarskynlikheids modelle

  1. 'n Houer bevat 1 wit, 6 rooi, 3 blou en 2 groen balle. 'n Bal word lukraak gekies. Wat is die waarskynlikheid dat dit die volgende kleur het:
    1. rooi
    2. blou of wit
    3. nie groen nie (wenk: dink 'komplement')
    4. nie groen of rooi nie?
  2. 'n Enkele kaart word lukraak uit 'n pak van 52 kaarte getrek. Wat is die waarskynlikheid dat die kaart:
    1. die 2 van harte is
    2. rooi
    3. 'n prent kaart
    4. 'n ase
    5. 'n nommer kaart kleiner as 4?
  3. Ewe getalle van 2-100 word elk op 'n kaart geskryf. Wat is die waarskynlikheid daarvan dat die getal'n veelvoud van 5 is as 'n kaart lukraak getrek word?

Waarskynlikheids identiteite

Die volgende resultate is van toepassing op waarskynlikhede, vir die steekproefruimte S en die twee gebeurtenisse A en B , binne S .

P ( S ) = 1
P ( A B ) = P ( A ) × P ( B )
P ( A B ) = P ( A ) + P ( B ) - P ( A B )

Ons kan die laaste resultaat demonstreer met behulp van 'n Venn-diagram. Die vereniging van A en B is die versameling van al die elemente in 'n of in B of beide.

Die waarskynlikheid dat gebeurtenis A voorkom word gegee as P(A) en die waarskynlikheid dat gebeurtenis B voorkom deur P(B) . Maar, as ons die sirkels wat hierdie gebeurtenisse voorstel ondersoek sal ons opmerk dat dat die waarskynlikheid 'n klein deeltjie van die ander gebeurtenis insluit. So gebeurtenis A bevat 'n stukkie van B en andersom. Dit word in die volgende diagram aangedui:

Ons merk op dat hierdie klein gedeelte die snyding van die twee gebeurtenisse is.

Wanneer ons die waarskynlikheid van P ( A B ) wil bepaal merk ons die volgende op:

  • Ons kan P ( A ) en P ( B ) bymekaar tel
  • Maar deur dit te doen tel ons die snyding dubbeld: een keer in P ( A ) en nog 'n keer in P ( B ) .
So, as ons net die waarskynlikheid van die snyding aftrek, dan sal ons vind dat die totale waarskynlikheid van die unie is: P ( A B ) = P ( A ) + P ( B ) - P ( A B )

Wat is die waarskynlikheid dat ons 'n swart of rooi kaart uit 'n pak van 52 kaarte sal trek.

  1. P ( S ) = n ( E ) n ( S ) = 52 52 = 1 want al die kaarte is of swart of rooi!

Wat is die waarskynlikheid om met 'n enkele trekbeurt uit 'n pak van 52 kaarte 'n klawer of ase te trek.

  1. P ( klawer ase ) = P ( klawer ) + P ( ase ) - P ( klawer ase )
  2. = 1 4 + 1 13 - 1 4 × 1 13 = 1 4 + 1 13 - 1 52 = 16 52 = 4 13

    Neem kennis hoe ons gebruik gemaak het van P ( C A ) = P ( C ) + P ( A ) - P ( C A ) .

Die volgende video verskaf 'n kort opsomming van sommige van die werk wat ons tot dusver behandel het.

Khan academy video oor waarskynlikheid

Oefeninge met waarskynlikheidsidentiteite

Beantwoord die volgende vrae:

  1. Rory oefen vir 'n skyfskiet kompetisie. Sy waarskynlikheid om die teiken te tref is 0,7 . Hy vuur vyf skote af. Wat is die waarskynlikheid dat al vyf skote mis is?
  2. 'n Boogskut skiet op 'n teiken. Die waarskynlikheid vir 'n kolskoot is 0,4 . As sy drie pyle skiet, wat is die waarskynlikheid van drie kolskote?
  3. 'n Dobbelsteen met die nommers 1,3,5,7,9,11 word gerol. Op dieselfde tyd word 'n ewekansige munt opgeskiet. Wat is die waarskynlikheid dat :
    1. Die munt land op "kop" en 'n 9 word gerol?
    2. Die munt land op "stert" en 'n 3 word gerol?
  4. Vier leerder skryf 'n toets. Die waarskynlikhede dat elkeen slaag is as volg. Sarah: 0,8 , Kosma: 0,5 , Heather: 0,6 , Wendy: 0,9 . Wat is die waarskynlikheid dat:
    1. al vier slaag?
    2. al vier druip?
  5. Met 'n enkele trekbeurt uit 'n pak van 52 kaarte, wat is die waarskynlikheid dat die kaart 'n ase of 'n swart kaart is?

Questions & Answers

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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