# Uniform distribution

1. May 4, 2013

### synkk

http://gyazo.com/b031e9d54f9512e5a583a4ed0ea28a0a

the answer is 2/3:

my attempt:
one side X~U[1,7]

the longer part of this side, call Y, where Y~[4,7]

P(Y>6) = 1/3

don't see how they got 2/3, they have 3 different methods in the answers, but none doing my method.

I'm thinking, as there are two sides in the rectangle and I've only limited mine to one side then if I multiply that by 2 I'll get the answer - or is that even relevant?

I'll give another example where I used my method and it works:

http://gyazo.com/a582f228a9dd22370303f25b69d5bf6c

part c)

the shorter side can be distributed by X~U[0,10]

P(X>8) = 1/5 which is the answer, but this doesn't work for the original method. Any ideas why? Thanks.

2. May 4, 2013

### FeDeX_LaTeX

I agree that X ~ U[1,7].

Where are you getting that Y ~ U[4,7]?

I would think that instead Y~U[3,9], where Y is the length of the other side. If one side of the rectangle is 1 cm, then the other side must be 9 cm. Similarly, if one side of the rectangle is 7 cm, the other side must be 3 cm. So Y ~ U[3,9]. Do you see where to go from here?

Is this an Edexcel S2 question?

3. May 4, 2013

### synkk

Yes it is

I got Y ~ U[4,7] where Y is the longer side, by setting Y as the longer side (not the other side). Median length is 4, so for Y to be the longer side, it has to be between [4,7].

I get the correct answer Y ~ U[3,9] where Y is the other side, and doing P(X>6) + P(Y>6), but I don't see why it doesn't work using the above method.

4. May 4, 2013

### FeDeX_LaTeX

I don't follow your reasoning. Your distribution doesn't represent Y to be the longer side. If Y ~ U[4,7], what happens when 5 < X < 7? Clearly, X is then the longer side.

5. May 4, 2013

### synkk

hm, I'm just confused now.

If they are asking for one side, and if Y ~ U[3,9] then why do we have to consider the other side (i.e X). Why can't we just consider Y

6. May 4, 2013

### FeDeX_LaTeX

They are asking what happens when the longer side is greater than 6 cm.

If X ~ U[1,7] and Y~U[3,9], then either side can be longer than 6 cm. We can have X > 6, or Y > 6. We can't just consider Y, because X can be larger than 6 cm also. Does that make sense?

7. May 4, 2013

### synkk

that makes sense

but if you see in the second example, I split up the whole length and I just considered the shorter side (which is what I tried to do when doing the mock), but it didn't work :\.

8. May 4, 2013

### FeDeX_LaTeX

You mean this?

I agree with the reasoning above. The shorter side must be between 0 and 10, otherwise it isn't the shorter side by definition.

9. May 4, 2013

### Ray Vickson

For the first question above: let Y be the OTHER (not longest) side and let Z = max(X,Y) be the length of the longest side. We have X+Y=10, so Y = 10 - X. We have Z = X if X > Y = 10-X, so Z = X if X > 5. Of course, Z = 10-X if X < 5. Draw a graph of the function Z(x) = max(x,10-x) for 4 <= x <= 7. What is the smallest value of Z(x)? What is the largest value? For what x-region do we have Z(x) > 6? What is the probability of that x-region?

BTW: I agree with your answer: P{Z > 6} = 1/3.

Last edited: May 4, 2013