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Draw a tree of possibilities

  1. Sep 24, 2011 #1
    hi can any one help my whit this questions :confused:

    Q1:

    Harry is an amateur birdwatcher who goes birdwatching in a region in which he knows there are only two species of thornbill – the buff-rumped thornbill and the yellow-rumped thornbill. Harry has some difficulty in differentiating the two species (but he does know a thornbill when he sees one). If Harry sees a buff-rumped thornbill, he will correctly identify it 80% of the time. If he sees a yellow-rumped thornbill, he will correctly identify it 60% of the time. In this region, 70% of the thornbills are buff-rumped.

    (i) If Harry sees a thornbill, what is the probability that he thinks it is a yellow-rumped?

    Harry goes birdwatching in this region with George who is a thornbill expert. If Harry correctly identifies a thornbill, George will correctly identify it 99% of the time. If Harry misidentifies a thornbill, George will correctly identify it 90% of the time.

    (ii) Harry and George both spot a thornbill which they both identify as a yellow-rumped. What is the probability that the bird is in fact buff-rumped?

    =============================================

    Q2:

    At the start of each month, Microsoft classifies Xbox Live users either as Gold, Silver or Bronze members. Gold members have a 70% chance of staying a Gold member, and can never become Bronze members by the end of the next month. If a user is a Silver member at the start of a given month, then they are 3 times as likely as becoming a Gold member compared with a Bronze member at the start of the next month. They also have a 40% chance of staying a Silver member.
    Finally, 50% of Bronze members remain Bronze members, while 50% become Silver members at the start of the next month. For the 50% of Bronze members who would become Silver members at the start of the next month, Microsoft randomly offers 10% of these users a chance to become a Gold member at the start of the next months (hence it is possible for Bronze members to become Gold members in one month). To accept this offer, the user must pay a small fee, so not everybody accepts. Microsoft estimates that 90% of Bronze members accept this offer.

    (a) Set up the probability transition matrix for this Markov chain, with a user’s membership considered at the beginning of each month.

    (b) If a user is a Bronze member at the beginning of April, what is the probability that they will be a Gold member at the beginning of June?

    (c) Determine the proportion of users who are Gold members in the long run.
     
  2. jcsd
  3. Sep 24, 2011 #2

    CompuChip

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    Re: Help :(

    It might be useful to draw a tree of possibilities:
    Code (Text):

     
             /\
            /  \
       yellow   not yellow
         /\          /\
        /  \        /  \
       /    \    right wrong
     right wrong
     
    Now add the given probabilities. Which path(s) through the tree identifyies(y) the situation in (i)?
     
  4. Sep 24, 2011 #3
    Re: Help :(

    what about Q1 (II) and Q2
     
  5. Sep 25, 2011 #4

    CompuChip

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    Re: Help :(

    You're welcome.

    Did you manage to do Q1 (I), because (II) is quite similar.

    Please be a little more specific in your question, we're not going to do your problems for you.
    Q2 is a really straightforward question about Markov chains (it's like they filled out "the" template for Markov chain problem sets :-)), so what are you having trouble with?
     
  6. Sep 29, 2011 #5
    Re: Help :(

    hi,

    Q1:

    Part 1:

    P(buff-rumped) = .7
    P(yellow-rumped) = .3

    P(Harry correctly identifies buff-rumped)=.8
    P(Harry incorrectly identifies buff-rumped)=.2


    P(Harry correctly identifies yellow-rumped)=.6
    P(Harry incorrectly identifies yellow-rumped)=.4

    P(Harry thinks it is a yelow-rumped)=.18

    hope this helps. If you truly understand this; you will do 2nd part without any difficulty, if not find a different subject that you will enjoy while you're learning!
     
  7. Sep 29, 2011 #6
    Re: Help :(

    Thanks,
    but i solve the problem this way and i don't know is it correct or not:

    i.
    p(yellow, harry thinks it's yellow) = 0.3 * 0.6 = 0.18
    p(yellow, harry thinks it's buff) = 0.3 * 0.4 = 0.12
    p(buff, harry thinks it's yellow) = 0.7 * 0.2 = 0.14
    p(buff, harry thinks it's buff) = 0.7 * 0.8 = 0.56
    => p(harry think it's yellow) = 0.18 + 0.14 = 0.32

    ii. There are two ways they can both say yellow:
    p(yellow, harry thinks yellow, george correctly agrees) = 0.18 * 0.99 = 0.1782
    p(buff, harry thinks yellow, george incorrectly agrees) = 0.14 * 0.10 = 0.014
    => p(actually buff) = 0.014 / (0.014 + 0.1782)
     
  8. Sep 29, 2011 #7

    CompuChip

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    Re: Help :(

    Your answer for i) looks good.
    For ii), why do you divide by (0.014 + 0.1782)? Isn't 0.014 already the answer?
     
  9. Sep 29, 2011 #8
    Re: Help :(

    you can see P ( A|B) = p(A [itex]\cap[/itex] B) /p (B)
     
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