التباين في الاحتمالات

47
3 ﺗﺼﻤﻴﻢ ﺍﻟﺪﺭﺱI II III IV V VI VII VIII

Transcript of التباين في الاحتمالات

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    .I

    .II

    .III

    .VI

    .V

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    ( + ) .IIV

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  • : .III

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  • : .IV

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    ( )ip X x =136

    236

    336

    436

    536

    636

    ix 8 9 10 11 12

    ( )ip X x =536

    436

    336

    236

    136

    1 2 3 4 5 6( ) 2 3 4 5 6 7

    36 36 36 36 36 365 4 3 2 1

    8 9 10 11 1236 36 36 36 36

    E X = + + + + + +

    + + + +

    ( )1( ) 2 6 12 20 30 42 40 36 30 22 1236

    E X = + + + + + + + + + +

    :( ) 7E X =

    ( p(X>6)p(X=12)

    1( 12)36

    p X = =

    ( 6)p X >( 6) ( 7) ( 8) ( 9)

    ( 10) ( 11) ( 12)p X p X p X p X

    p X p X p X > = = + = + = +

    = + = + =

    6 5 4 3 2 1( 6) 7 8 9 10 11 1236 36 36 36 36 36

    p X > = + + + + +

    :182( 6)252

    p X > =

  • 123456

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    .2

    = pi1 pi0 m

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    i ii

    m x P V =

    - =

    = + + - = = X p2 ,0 )3,0 2 ,0 3,0( 1 )0 (:

    = V22 = m 1

    = + + + - = = X p50,0 )1,0 51,0 2 ,0 5,0( 1 )001 (:

    = V017 = m 03

    = + + + + - = = X p51,0 )50 ,0 1,0 1,0 4 ,0 2,0( 1 )6 (:

    = V54,11 = m 5,0

    .3

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    M

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  • 10.

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    0,8 0,3 0,86 0, 24

    P A B P A P B P A B = + -

    = + - =

    ( ) ( ) ( )P A B P A P B = AB .11.

    ( )P A B ( )BP A ( )AP B

    ( ) 0,3P A = ( ) 0,7P B = ( ) 0,8P A B =

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    P A B P A P B P A B = + -

    = + - =

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    P A BP A

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    P A

    = = =

  • : .IIV

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    - = = B A p B A p B A p) ( 1 ) () ( 8,0( .4

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    = + = 9,0 18,0 90,0

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  • : .IIIV

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    E T E S *

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    = T p

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    .

    . .2

    ( .3

    .

    . = B A: BA(

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