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zarkam21

  • one year ago

Help please

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  1. linn99123
    • one year ago
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    with?

  2. zarkam21
    • one year ago
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  3. Michele_Laino
    • one year ago
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    each time 2 outcomes are possible, namely head and tail

  4. zarkam21
    • one year ago
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    so 8 outcomes right?

  5. Michele_Laino
    • one year ago
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    yes!

  6. zarkam21
    • one year ago
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    so for part one, it would be 2*2*2*2=8?

  7. zarkam21
    • one year ago
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    now for part 2

  8. Michele_Laino
    • one year ago
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    please wait: it is 2*2*2*2=16

  9. zarkam21
    • one year ago
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    yeah but isn't it a total of 8 outcomes

  10. Michele_Laino
    • one year ago
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    more precisely: we have 2 possible outcomes for each flip, so when we multiply, we get: \[\Large 2 \times 2 \times 2 \times 2 = 16\] that number expresses the possible outcomes

  11. Michele_Laino
    • one year ago
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    if we add: 2+2+2+2 we get 8 nevertheless that number is not interesting for your exercise

  12. zarkam21
    • one year ago
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    okay so 2*2*2*2=16 :]

  13. Michele_Laino
    • one year ago
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    correct! I'm sorry for my previous error

  14. zarkam21
    • one year ago
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    It is alright!

  15. zarkam21
    • one year ago
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    Let's move on to part II

  16. Michele_Laino
    • one year ago
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    ok!

  17. Michele_Laino
    • one year ago
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    here we have to do the subsequent reasoning: lets suppose to start from a head (H) then the next flip can be another head or a tail, so we have these outcomes: \[\Large \begin{gathered} HH \hfill \\ HT \hfill \\ \end{gathered} \]

  18. zarkam21
    • one year ago
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    Yes

  19. Michele_Laino
    • one year ago
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    whereas if we start from a tail (T), then next flip can be another tail or a head, so we get these other possible outcomes: \[\Large \begin{gathered} TH \hfill \\ TT \hfill \\ \end{gathered} \]

  20. zarkam21
    • one year ago
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    okay

  21. Michele_Laino
    • one year ago
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    in total we got these four outcomes: \[\Large \begin{gathered} 1)HH \hfill \\ 2)HT \hfill \\ 3)TH \hfill \\ 4)TT \hfill \\ \end{gathered} \]

  22. zarkam21
    • one year ago
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    So that is part II? HH,HT, TH, and TT

  23. Michele_Laino
    • one year ago
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    no, we have to finish :)

  24. zarkam21
    • one year ago
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    okay

  25. Michele_Laino
    • one year ago
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    next, let's start from the outcome 1) then next flip can be a head or a tail, so we get these outcomes: \[\Large \begin{gathered} HHH \hfill \\ HHT \hfill \\ \end{gathered} \]

  26. zarkam21
    • one year ago
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    okay

  27. Michele_Laino
    • one year ago
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    now, let's start from outcome 2), then next flip can be a head ot a tail, so we get these new outcomes: \[\Large \begin{gathered} HTH \hfill \\ HTT \hfill \\ \end{gathered} \]

  28. zarkam21
    • one year ago
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    Yes, :} Would the next be, THH and THT

  29. Michele_Laino
    • one year ago
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    correct!

  30. zarkam21
    • one year ago
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    and I'm guessing the next is TTH and TTT

  31. Michele_Laino
    • one year ago
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    and the next two new outcomes are: \[\Large \begin{gathered} TTT \hfill \\ TTH \hfill \\ \end{gathered} \] since I have started from outcome 4)

  32. zarkam21
    • one year ago
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    Does it matter what order?

  33. Michele_Laino
    • one year ago
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    no, the order of the outcomes, is not important so, in total we got these outcomes: \[\Large \begin{gathered} a)HHH \hfill \\ b)HHT \hfill \\ c)HTH \hfill \\ d)HTT \hfill \\ e)THT \hfill \\ f)THH \hfill \\ g)TTT \hfill \\ h)TTH \hfill \\ \end{gathered} \]

  34. zarkam21
    • one year ago
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    Okay :}

  35. Michele_Laino
    • one year ago
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    now, we have to start from outcome a) and, with the same procedure we get these two new outcomes: \[\Large \begin{gathered} HHHH \hfill \\ HHHT \hfill \\ \end{gathered} \]

  36. zarkam21
    • one year ago
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    Yup :}

  37. zarkam21
    • one year ago
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    and then it would be HHTH and HHTT

  38. Michele_Laino
    • one year ago
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    starting from outcome b) I get these ones: \[\Large \begin{gathered} HHTH \hfill \\ HHTT \hfill \\ \end{gathered} \]

  39. Michele_Laino
    • one year ago
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    and we have to repeat the same procedure for the remaining outcomes c), d), e), f), g) and H)

  40. Michele_Laino
    • one year ago
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    so we get these outcomes: \[\Large \begin{gathered} 1)HHHH \hfill \\ 2)HHHT \hfill \\ 3)HHTH \hfill \\ 4)HHTT \hfill \\ 5)HTHH \hfill \\ 6)HTHT \hfill \\ 7)HTTT \hfill \\ 8)HTTH \hfill \\ 9)THTT \hfill \\ 10)THTH \hfill \\ 11)THHT \hfill \\ 12)THHH \hfill \\ 13)TTTT \hfill \\ 14)TTTH \hfill \\ 15)TTHT \hfill \\ 16)TTHH \hfill \\ \end{gathered} \]

  41. Michele_Laino
    • one year ago
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    that is the answer for part II

  42. Michele_Laino
    • one year ago
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    as you can see there are 16 possible outcomes, namely the number which we have computed in part I

  43. zarkam21
    • one year ago
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    Yes :}

  44. Michele_Laino
    • one year ago
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    now, for part III, we have highlight those outcomes which contains two T, namely TT

  45. Michele_Laino
    • one year ago
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    how many outcomes contain two (T)

  46. zarkam21
    • one year ago
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    2 or more or just two

  47. Michele_Laino
    • one year ago
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    I got six outcomes, precisely the subsequent outcomes: #4, #6, #8 ,#10, #11, #16

  48. zarkam21
    • one year ago
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    Okay :}

  49. Michele_Laino
    • one year ago
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    so the probability to have two flips which comes up with (T) is: p=favorable outcomes/possible outcomes=6/16 am I right?

  50. zarkam21
    • one year ago
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    Yes

  51. zarkam21
    • one year ago
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    But would't 6/16 be reduced

  52. zarkam21
    • one year ago
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    @Michele_Laino

  53. Michele_Laino
    • one year ago
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    it is: \[\Large p = \frac{6}{{16}} = \frac{3}{8}\]

  54. zarkam21
    • one year ago
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    for part II right?

  55. Michele_Laino
    • one year ago
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    for part III is right!

  56. zarkam21
    • one year ago
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    oh okay part III?

  57. zarkam21
    • one year ago
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    next is part IV

  58. Michele_Laino
    • one year ago
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    ok!

  59. Michele_Laino
    • one year ago
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    here we have to highlight those outcomes which contain three (H). How many outcomes do contain three (H) ?

  60. zarkam21
    • one year ago
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    Okay so #2, #3, #5, and #12

  61. Michele_Laino
    • one year ago
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    correct! we have 5 outcomes, so the requested probability, is: p=favorable outcomes/possible outcomes=\[\Large p = \frac{5}{{16}}\]

  62. Michele_Laino
    • one year ago
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    oops.. sorry we have 4 outcomes, so the requested probability is: \[\Large p = \frac{4}{{16}} = \frac{1}{4}\]

  63. zarkam21
    • one year ago
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    okay :}

  64. zarkam21
    • one year ago
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    Thank you sooo much !!!!!!!

  65. Michele_Laino
    • one year ago
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    :) :)

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