# Thread: ACST212 - Combinatorial Probability

1. ## 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

2. ## Re: ACST212 - Combinatorial Probability

Originally Posted by BenHowe
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.

3. ## Re: ACST212 - Combinatorial Probability

Ok cool thanks for that. I managed to do the rest of the q, thanks heaps

4. ## 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

$\frac{1}{3^{n-1}}[^{n+2}C_{3}+^{n+5}C_{6}+...+^{2n-1}C_{n+1}][1-^{n}C_{1}+^{n}C_{2}-...+^{n}C_{n}]$

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

5. ## 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?

6. ## Re: ACST212 - Combinatorial Probability

Originally Posted by BenHowe
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.

7. ## Re: ACST212 - Combinatorial Probability

Sorry would you be able to elaborate, like specifically how the cases from 100p5 collapse to 5!?

8. ## 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 ?

9. ## Re: ACST212 - Combinatorial Probability

Originally Posted by Bartimaeus
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.

10. ## 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

11. ## Re: ACST212 - Combinatorial Probability

Originally Posted by Bartimaeus
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.

12. ## 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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