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Thread: ACST212 - Combinatorial Probability

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    ACST212 - Combinatorial Probability

    Hey need some help with questions. Here's the 1st one.

    Capture.PNG

    Edit I'd like a hint not the answer. Thanks heaps
    2nd Year Applied Finance/Actuarial Studies @ MQ
    GPA 6.77 WAM 87%
    Part 1's: CT1,CT2, 1/2 CT3,CT7

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    Re: ACST212 - Combinatorial Probability

    Quote Originally Posted by BenHowe View Post
    Hey need some help with questions. Here's the 1st one.

    Capture.PNG

    Edit I'd like a hint not the answer. Thanks heaps
    Try and consider the die modulo 3.
    BenHowe likes this.

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    Re: ACST212 - Combinatorial Probability

    Ok cool thanks for that. I managed to do the rest of the q, thanks heaps
    2nd Year Applied Finance/Actuarial Studies @ MQ
    GPA 6.77 WAM 87%
    Part 1's: CT1,CT2, 1/2 CT3,CT7

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    Re: ACST212 - Combinatorial Probability

    Edit: I was wrong xD

    Hey here's the rest of the q Capture.PNG and I managed to do b) but my expression with c inst quite working.

    I get



    Although I know I should be able to show the answer is always 1/3 because say for the k-1 th roll (k is a positive finite integer) you can think about S mod 3 (where S is the sum of the previous remainders) and it will either be 0,1,2. For each of these outcomes there are exactly 2 corresponding numbers that can be rolled and hence k mod n will be 0. So it will be 2/6=1/3
    2nd Year Applied Finance/Actuarial Studies @ MQ
    GPA 6.77 WAM 87%
    Part 1's: CT1,CT2, 1/2 CT3,CT7

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    Re: ACST212 - Combinatorial Probability

    Dw I actually got it this time.

    Here's the next question

    Capture.PNG

    I've done a,b and some of c but I'm struggling to show the cases are equally likely. any hints?
    2nd Year Applied Finance/Actuarial Studies @ MQ
    GPA 6.77 WAM 87%
    Part 1's: CT1,CT2, 1/2 CT3,CT7

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    Re: ACST212 - Combinatorial Probability

    Quote Originally Posted by BenHowe View Post
    Dw I actually got it this time.

    Here's the next question

    Capture.PNG

    I've done a,b and some of c but I'm struggling to show the cases are equally likely. any hints?
    They're equally likely by symmetry.

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    Re: ACST212 - Combinatorial Probability

    Sorry would you be able to elaborate, like specifically how the cases from 100p5 collapse to 5!?
    2nd Year Applied Finance/Actuarial Studies @ MQ
    GPA 6.77 WAM 87%
    Part 1's: CT1,CT2, 1/2 CT3,CT7

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    Re: ACST212 - Combinatorial Probability

    Hi I have a coupon collector problem, but instead of the expected number of attempts to get all the different coupons, what's the expected number of a particular coupon in satisfying the above constraint?

    The actual questions is:

    There is an urn with a large number of blue and red balls, what is the expected number of blue balls such that we must have at least 1 blue and 1 red ball ?

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    Re: ACST212 - Combinatorial Probability

    Quote Originally Posted by Bartimaeus View Post
    Hi I have a coupon collector problem, but instead of the expected number of attempts to get all the different coupons, what's the expected number of a particular coupon in satisfying the above constraint?

    The actual questions is:

    There is an urn with a large number of blue and red balls, what is the expected number of blue balls such that we must have at least 1 blue and 1 red ball ?
    It's half the answer to the usual coupon collector problem (which is to find the expected number of draws to get at least one of every coupon).

    In general, it'd be 1/K times the answer to the usual problem, where K is the no. of colours (coupons) available.

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    Re: ACST212 - Combinatorial Probability

    Thanks heaps for the response. What is the best way of showing this? Like conditional expectation theorem? I'm just wondering because I need to be able to clearly show correct working for q's such as this in the exam.

    I can do it by symmetry luike youve said above, but yeah I need to be able to do by as many methods as possible
    Last edited by Bartimaeus; 11 Nov 2017 at 8:55 AM.

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    Re: ACST212 - Combinatorial Probability

    Quote Originally Posted by Bartimaeus View Post
    Thanks heaps for the response. What is the best way of showing this? Like conditional expectation theorem? I'm just wondering because I need to be able to clearly show correct working for q's such as this in the exam.

    I can do it by symmetry luike youve said above, but yeah I need to be able to do by as many methods as possible
    Probably best to use symmetry. Don't need to use conditional expectation. But for the particular question you asked (with blue and red balls), it's also easy to do it by conditioning on the first draw's colour, and noting the number of draws to get the remaining colour after that is just Geometrically distributed.

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    Re: ACST212 - Combinatorial Probability

    So you would proceed as :

    Let R be the random variable denoting the number of red balls selected
    Let R1 be the event, the first ball selected is red
    Let B1 be the event, the first ball selected is blue

    E(R)=P(R1)E(R+1|R1)+P(B1)E(R|B1), by the Conditional Expectation Theorem

    Now consider E(R+1|R1), we have obtained 1 red ball. So we will continue to select balls until we have obtained 1 blue ball. So R+1 is a geometric random variable with parameter 1/2. So E(R+1|R1)=2.

    Now consider E(R|B1), we have obtained 1 blue ball. We only need 1 red ball. So E(R|B1)=1

    Thus E(R)=1/2 X [2] +1/2 X [1]

    =3/2 as required

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