# Fetch Minimal Record with Probability

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.

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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

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 $1$.

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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