Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have three records. the records means intervals,

  1. $A: [1, 5]$

  2. $B: [2, 6]$

  3. $C: [4, 6]$

A, B and C are three humidity sensors. Value of A is between 1 to 5, B is between 2 to 6, and C is between 4 to 6

Assume the humidity data is uniform distributed(uniform distribution). I'd like to know which sensor would give the minimal value(lowest humidity), with probability.

share|improve this question
3  
You might want to elaborate slightly. –  copper.hat Nov 2 '12 at 7:29
    
Are those points or intervals? What does "minimal data" mean? You tagged this with "uniform-distribution" but have not mentioned that in the question. –  Henry Nov 2 '12 at 8:06
    
@Henry I've edited the question –  fengd Nov 2 '12 at 13:43
add comment

1 Answer

up vote 0 down vote accepted

Clearly the minimum falls in the ranges $[1,2)$ or $[2,4)$ or $[4,5]$ so the probabilities A, B or C give the minimum respectively are

$$\int_{a=1}^2 \frac{da}{5-1}\cdot1\cdot1 + \int_{a=2}^4 \frac{da}{5-1} \cdot \frac{6-a}{6-2} \cdot 1 + \int_{a=4}^5 \frac{da}{5-1}\cdot \frac{6-a}{6-2}\cdot\frac{6-a}{6-4}$$

$$0 + \int_{b=2}^4 \frac{5-b}{5-1}\cdot \frac{db}{6-2}\cdot 1 + \int_{b=4}^5\cdot \frac{5-b}{5-1} \frac{db}{6-2}\cdot\frac{6-b}{6-4}$$

$$0 + 0 + \int_{c=4}^5 \frac{5-c}{5-1}\cdot \frac{6-c}{6-2}\cdot\frac{dc}{6-4}$$

and if you calculate these they should add up to $1$.

share|improve this answer
    
since they are all uniform distribution, I think I can get the value of these expressions, right? –  fengd Nov 3 '12 at 7:57
    
@Jun1st: One of the integrals is of a constant, two of linear function and three of a quadratic function. –  Henry Nov 3 '12 at 9:47
    
to the probability of A between [2, 4), I understand the da/(5-1), but not (6-a)/(6-2), why? on my understanding, A has 50% chance between [2, 4), B has 50% change between [2,4). they are independent, so either one has 25% chance to be the minimal value –  fengd Nov 3 '12 at 14:55
    
If $A=a$ in $[2,4)$ then the probability $B$ is greater is $\dfrac{6-a}{6-2}$ –  Henry Nov 3 '12 at 16:04
    
I think the probability for B is greater is (6-4)/(6-2) –  fengd Nov 4 '12 at 5:16
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.