1 Attachment(s)

Probability Independent Events

Hi guys,

I'm stuck on these 2 questions, but I think that once I can do the first, the other should be reasonably straightforward.

I would really appreciate some help.

I've attached a png if it's easier to read.

Thanks!

Quote:

A genetic experiment on cell division can give rise to at most 2k cells,

and the probability function describing the variation in the number of

cells recorded is assumed to be

p(x) = x(1 -theta)/1-theta^(2k+1)

x = 0, 1, 2, ..., 2k

where 0 < theta < 1:

(b) Two independent replicates of the experiment are performed. What

are the probabilities

ii. that the number of cells produced in the two replicates are

the same;

iii. That the total number of cells produced from the two replicates

is exactly y, for 0 y 2k?

Re: Probability Independent Events

Hey UncleFester.

Hint: If you have two independent events A and B then the likelihood of both happening is P(A and B) = P(A)*P(B). If you have probability functions for each event, the multiply them to get the joint density function.

Re: Probability Independent Events

Hi chiro,

I was thinking I need to sum the square of p(x) for each value of x then simplify it somehow to remove x from the equation. Is that right? If so, do you have any thoughts on how to simplify it?

Re: Probability Independent Events

Based on what I said above, can you describe the events A and B and the value of the probability involved?

Re: Probability Independent Events

A is the first attempt at the experiment, B is the second attempt. Both have the same probabilities for any particular value of x (as in OP)

Re: Probability Independent Events

Could you please be more specific? Can you give a mathematical expression for the probabilities please? (Making your attempts specific will help us see your method of thinking and where you went wrong or right and this procedure is encouraged here on these forums).

1 Attachment(s)

Re: Probability Independent Events

Hi,

The way I like to think about such problems is with independent random variables. If you don't know about random variables, I guess this post is not much help. You were exactly right in your response 3; even if you are hazy about r.v.'s, the following attachment shows a simplification of the expression as you asked. Furthermore, the 2nd part is briefly addressed.

Attachment 29856

Re: Probability Independent Events

Thanks, I'm pretty sure I got it. It's handed in now so time will tell :D