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mathmath333

  • one year ago

Probablity question

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  1. mathmath333
    • one year ago
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    que: On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits (i) A before B? (ii) A before B and B before C? (iii) A first and B last? (iv) A either first or second? (v) A just before B?

  2. anonymous
    • one year ago
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    probably not

  3. mathmath333
    • one year ago
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    The number of ways by which veena can visit is 4!=24

  4. welshfella
    • one year ago
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    yes

  5. mathmath333
    • one year ago
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    i can'r go further

  6. mathmath333
    • one year ago
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    *can't

  7. welshfella
    • one year ago
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    i guess you can list ways in which a comes before B and count them A bit tedious though

  8. mathmath333
    • one year ago
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    yes that is tediuos i want another time saving way

  9. welshfella
    • one year ago
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    or if A is the first then how many ways can the others be visited 3*2 = 6 if A is 2nd and B 3rd or 4th - thats 2 ways

  10. welshfella
    • one year ago
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    - no the last one is 4 ways then we have A 3rd and B 4th - 2 ways total 6 + 4 + 2 = 12 so first one is 12/24 = 1/2

  11. welshfella
    • one year ago
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    to be honest I hate probability!

  12. mathmath333
    • one year ago
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    sumilarly B before C = 12 ways ?

  13. welshfella
    • one year ago
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    yes must be

  14. welshfella
    • one year ago
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    can we multiply these now to get the compound probability?

  15. mathmath333
    • one year ago
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    ok i need to find now the common between A before B and B before C

  16. welshfella
    • one year ago
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    the question is are they independent events?

  17. mathmath333
    • one year ago
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    i don't understand how independent events means here

  18. mathmath333
    • one year ago
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    there are 12 events in both A before B and B before C and i need to find the common events between them

  19. mathmath333
    • one year ago
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    (A before B) \(\Large \cap\) (B before C)

  20. welshfella
    • one year ago
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    by listing and counting there are 4 of them

  21. mathmath333
    • one year ago
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    ok so i need to list them necessrily

  22. welshfella
    • one year ago
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    i'm afraid i dont remember all the set theory applied to probability

  23. mathmath333
    • one year ago
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    ok so same for 3, 4 and 5 , listing ?

  24. welshfella
    • one year ago
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    questions iii to v are quite straightforward - i think!!

  25. welshfella
    • one year ago
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    for q iii it is the number ow ways you can arrange the 3 letters between A and B

  26. welshfella
    • one year ago
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    = 6 right?

  27. mathmath333
    • one year ago
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    3!=6 , yes

  28. welshfella
    • one year ago
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    for iv A first = 3! ways + 31 ways with A second

  29. welshfella
    • one year ago
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    * 3!

  30. mathmath333
    • one year ago
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    ok

  31. welshfella
    • one year ago
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    v AB to begin - - 2 ways AB 2nd and 3rd - 2 waya AB last - 2 ways

  32. anonymous
    • one year ago
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    nice question

  33. welshfella
    • one year ago
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    yea it is

  34. mathmath333
    • one year ago
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    AB-- A_B__ A__B _AB__ _A_B_ _A__B __AB i calculate 7 ways for v.)

  35. mathmath333
    • one year ago
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    @welshfella

  36. anonymous
    • one year ago
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    |dw:1439460306887:dw| just another way of thinking about it"!

  37. mathmath333
    • one year ago
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    is the 7 ways correct for last one ?

  38. welshfella
    • one year ago
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    No i dont think so - isn't it A just before B (A and B together?)

  39. phi
    • one year ago
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    ***(i) A before B? *** I guess we could count the different routes but it seems there is nothing special about A and B, and half the time A will be first, and B second (and vice versa)

  40. Zarkon
    • one year ago
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    for (v) A just before B? put A and B next to each other and count it as one item let X=AB then we want the number of arrangements of XCD which is 3!=6 the probability easily follows

  41. phi
    • one year ago
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    (ii) A before B and B before C? it seems difficult to find the prob of each event and then subtract off their intersection. Rather, how about writing A B C and asking, where can D go? we can list DABC, ADBC, ABDC, ABCD so 4 patterns out of 24 and prob 4/24= 1/6

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