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hockeychick23

  • one year ago

Using the tree diagram, you just constructed and make a table showing the probability distribution of getting zero, one, or both questions right by guessing. (i attached the tree diagram)

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  1. hockeychick23
    • one year ago
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  2. hockeychick23
    • one year ago
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    @welshfella @kropot72

  3. hockeychick23
    • one year ago
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    I got: Probability of getting zero questions right by guessing: .5626(.75)= .42195 Probability of getting one question right by guessing: (0.75)(.1875)= .140625 (0.25)(.0625)= 0.015625 .140625+0.015625= .15625 but I'm not sure if i did it correctly, can someone check my answer please?

  4. UnkleRhaukus
    • one year ago
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    don't multiply

  5. UnkleRhaukus
    • one year ago
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    the number on the right most nodes are the total probabilities already

  6. hockeychick23
    • one year ago
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    oh ok so the probability of getting zero right is .5625, getting one right is .25+.1875=.4375 and getting both right would be .1875?

  7. UnkleRhaukus
    • one year ago
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    |dw:1433151671567:dw|

  8. UnkleRhaukus
    • one year ago
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    yes the probability of getting zero right is .5625,

  9. UnkleRhaukus
    • one year ago
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    the probability of getting 1 right is the sum of the two probability of the nodes in the tree that are exactly 1 right ansewer

  10. hockeychick23
    • one year ago
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    yea i got .25+.1875=.4375

  11. UnkleRhaukus
    • one year ago
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    ie P(1) = P(wr) + P(rw)

  12. UnkleRhaukus
    • one year ago
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    0.25 is the probability that the first question is right, but we want to add the probabilities that exactly one question is right

  13. hockeychick23
    • one year ago
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    oh sorry so its .1875+.0625= .25

  14. UnkleRhaukus
    • one year ago
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    nope

  15. UnkleRhaukus
    • one year ago
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    |dw:1433152150638:dw|add these

  16. hockeychick23
    • one year ago
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    .375

  17. UnkleRhaukus
    • one year ago
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    yes

  18. hockeychick23
    • one year ago
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    and then both right would be .1875

  19. UnkleRhaukus
    • one year ago
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    **error in diagram |dw:1433152335528:dw|

  20. UnkleRhaukus
    • one year ago
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    there is only one leaf on the tree that has both right answers P(rr) =

  21. hockeychick23
    • one year ago
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    yea that leaf says .0625

  22. UnkleRhaukus
    • one year ago
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    |dw:1433152413453:dw|

  23. UnkleRhaukus
    • one year ago
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    yes,

  24. hockeychick23
    • one year ago
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    i think i drew my diagram wrong when i constructed it

  25. UnkleRhaukus
    • one year ago
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    why do you say that? it seems right to me its multiple choice (four options) two questions

  26. hockeychick23
    • one year ago
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    oh i just flip flopped the wrong and right on the second leaf

  27. UnkleRhaukus
    • one year ago
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    ah yeah

  28. UnkleRhaukus
    • one year ago
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    if we add up all the probabilities P(0) = 0.5625 P(1) = 0.1875+0.1875 = 0.3750 P(2) = 0.0625 P(0)+P(1)+P(2) = 1

  29. hockeychick23
    • one year ago
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    oh awesome! so if i made it into a table like it was asking could i just draw it like this:

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  30. UnkleRhaukus
    • one year ago
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    wcheck that last entry in the table again

  31. hockeychick23
    • one year ago
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    oh its not .1875? I thought that was what it was

  32. hockeychick23
    • one year ago
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    oh its .375

  33. UnkleRhaukus
    • one year ago
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    for both right: 1/4 * 1/4 =

  34. UnkleRhaukus
    • one year ago
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    nope 0.375 is for exactly 1 right

  35. hockeychick23
    • one year ago
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  36. hockeychick23
    • one year ago
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    sorry i typed it incorrectly, this is the table

  37. UnkleRhaukus
    • one year ago
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    now to be sure check the sum all the probabilities .5625 + .375 + .0625 =

  38. hockeychick23
    • one year ago
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    .5625+.375+0.0625=1

  39. UnkleRhaukus
    • one year ago
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    Yay!

  40. hockeychick23
    • one year ago
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    :) thanks so much!!

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