# Thread: General Probability Problem (Actuary question)

1. ## General Probability Problem (Actuary question)

https://imgur.com/a/nMeLwip

Not sure what to do with this problem:

I know that the difference is the same for p_(0)-p_(1)=p_(1)-p_(2)=p_(2)-p_(3)=p_(3)-p_(4)=p_(4)-p_(5)

I know we are looking for p_(4)+p_(5)

2. ## Re: General Probability Problem (Actuary question) Originally Posted by math951 https://imgur.com/a/nMeLwip
Not sure what to do with this problem:
I know that the difference is the same for p_(0)-p_(1)=p_(1)-p_(2)=p_(2)-p_(3)=p_(3)-p_(4)=p_(4)-p_(5)
I know we are looking for p_(4)+p_(5)
Do you know that $\displaystyle \sum\limits_{k = 0}^5 {{p_k}} = 1~?$ If so explain.

If $\delta=p_0-p_1$ is it true that $p_n=\delta+p_{n+1},~\;n=1,2,3,4~?$ Can you find $\delta~?$

3. ## Re: General Probability Problem (Actuary question)

I would approach it as an arithmetic progression where a is P(0) and the common difference is d.

So the terms are a, a+d, a+2d, a+3d, a+4d, a+5d.

All of these add to 1 (total of probabilities). The first 2 add to 0.4. This will give 2 simultaneous equations. Then find sum of last 2.

4. ## Re: General Probability Problem (Actuary question) Originally Posted by Plato Do you know that $\displaystyle \sum\limits_{k = 0}^5 {{p_k}} = 1~?$ If so explain.

If $\delta=p_0-p_1$ is it true that $p_n=\delta+p_{n+1},~\;n=1,2,3,4~?$ Can you find $\delta~?$
Doesn't the probability space always have to equal one?

5. ## Re: General Probability Problem (Actuary question) Originally Posted by Debsta I would approach it as an arithmetic progression where a is P(0) and the common difference is d.

So the terms are a, a+d, a+2d, a+3d, a+4d, a+5d.

All of these add to 1 (total of probabilities). The first 2 add to 0.4. This will give 2 simultaneous equations. Then find sum of last 2.
You have the progression backwards.

$$p_0=p_5+5d, p_1=p_5+4d, p_2=p_5+3d, p_3=p_5+2d, p_4=p_5+d, p_5=p_5$$

6. ## Re: General Probability Problem (Actuary question) Originally Posted by SlipEternal You have the progression backwards.

$$p_0=p_5+5d, p_1=p_5+4d, p_2=p_5+3d, p_3=p_5+2d, p_4=p_5+d, p_5=p_5$$
Not really, it's just that my d will be negative.

7. ## Re: General Probability Problem (Actuary question) Originally Posted by Debsta Not really, it's just that my d will be negative.
Oh, true. My bad.