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    I can do (i) and (ii) but stuck on (iii) and (iv)

    Every year 2 teams, the ramblers and the strollers, meet for a quiz night. From past results it seems that in years where the ramblers win, prob of them winning next year is 0.7 and in years when strollers win, prob of them winning is 0.5. It is not possible for the quiz to result in a tie.

    The ramblers won the quiz in 1996.
    (i) Draw a probability tree for the three years t0 1999.
    I drew tree correctly - no problem.

    (ii) Find the probability that the strollers will win in 1999.
    I added up all probs - eg 0.147+0.105+0,045+0.075=0.372 - that was correct answer.

    (iii) If the strollers win in 1999, what is the probability that it will be their first win for at least three years?

    I just selected the RRS option on tree above - 0.147. But this was incorrect. The correct answer apparently is 0.395. But how do you get that?

    (iv) Assuming that the strollers win in 1999, find the smallest value of n such that the probability of the ramblers winning the quiz for n consecutive years after 1999 is less than 5%.

    For this one I did this:
    0.5^x < 0.05
    log(0.05)/log(0.5) = 4.32., therefore n = 5. But that was incorrect, answer was 8. But how?

    Angus
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    For iii), you need to think in a slightly different way. You can certainly find a leaf of the tree which represents the Strollers winning in 1999, after losing every previous year given. But you are given already that they've won, so the probability of losing every previous year is affected (I know it sounds weird, but trust me, it works :P). In fact, you need to do: \frac{P(Strollers-win-for-the-first-time-in-'99)}{P(Strollers-win-AT-ALL-in-'99)}

    So add up all of the leaves in which the Strollers win in '99, and do 0.147 over that.

    As for iv), you've forgotten that if the Ramblers win the probability of them winning the next year goes up; you need to take that into account.
 
 
 
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Updated: April 12, 2011
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