# How to find probability relation

I'm not really sure what the term would be for such a problem (is it probability distribution?) but here it is.

I have a list divided into 4 sections and I am looking for a particular item.

The probability that the item is in the first fourth of the list is 3 times that of the second fourth of the list.

The probability that the item is in the last fourth of the list is twice that of the second fourth of the list.

And the probability that the item is in the third fourth of the list is twice the probability it is in the last fourth of the list.

Giving a relation like: [3x | x y | 2z | 2y z]

How can I find the probability that the item is in a particular part of the list? ie What is the probability it is in the first fourth of the list?

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Since I am not fond of fractions, it is numerically convenient to let $x$ be the probability the item is in the second quartile.

The probability it is in the first quartile is $3x$.

The probability it is in the last is $2x$.

The probability it is in the third is $2(2x)$, that is, $4x$.

So $x+3x+2x+4x=1$. Now we know everything.

Remark: If we like lots of variables, and subscripts, we can let $p_1$ be the probability the item is in the first quartile, $p_2$ the probability it is in the second, and so on.

The given information can be translated into equations, such as $p_1=3p_2$, and so on. We also have $p_1+p_2+p_3+p_4=1$. Solve.

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I'm not quite following you, if the probability for the 3rd quartile is 4x and for the first is 3x, then isn't that saying there is a larger chance it is in the 3rd quartile? That may be correct, but the way the problem is worded it sounds like the largest chance is the first quartile. – schwiz Feb 7 '13 at 6:41
It says that probability of last fourth is twice the second, and that third fourth is twice last fourth, so third fourth is $2(2x)$. – André Nicolas Feb 7 '13 at 6:44
Ok this definitely helps, I think the wording of the question is bothering me but I get the concept. I'll accept your answer once I get it worked out in case I have more questions or others want to chime in. The X-route just confuses me and while working out with the Ps I didn't get the sum to equal 1 when I was finished so I think its the wording for sure that is throwing me off. – schwiz Feb 7 '13 at 7:11
The simple way is the one I described first. For the quartiles, we get in order $3/10$, $1/10$, $4.10$, and $2/10$. If you want to use $p_i$, or $w,x,y,z$ we get $p_1=3p_2$ (I had a typo), $p_4=2p_2$, $p_3=2p_4$, $p_1+\cdots+p_4=1$. – André Nicolas Feb 7 '13 at 7:15
Ok, yeah so that is where I was getting confused, when using x's q1 is x and q2 is 3x, but with p p1 = 3p2, it switches around and that threw me off. But I think I understand it better using the p even though the x notation is more desirable. – schwiz Feb 7 '13 at 7:22