Probability Problem - Random variable

#1
A company wants to improve waste management in its premises . We put
a yellow bin for recovering recyclable materials , a brown to recover compostable materials and another black for waste. Analysis showed that recovered in
yellow tray random volume in the interval [ B, B + D] liters. This amount is in
the interval [ D, B + D] liters for brown container and is in the interval [ C, C + D] to liters
black tray. It is assumed that the quantities recovered different bins are mutually
independent.
Since the volume of the yellow bin is greater than that of brown , which is the
probability that the quantity in the black bin is greater than the sum of the two
other trays ?



I try to solve this problem but the thing is, all i have in mind is continued random variable. I dont know where to start. My randomb variable should be the volume i guess ?
 
#2
Yes, the random variable for each bin is the volume of material dumped into each bin. In the absence of relevant information, it makes sense to assume that these volumes are uniformly distributed.

The problem then becomes one of simple proportions, as illustrated in the attached image. You can also derive the restrictions on the values of B, C and D in relation to one another, given that a real-world physical volume must be ≥ 0, and a probability p is always limited to 0 ≤ p ≤ 1.

The info herein should be enough for you to solve the problem.
 
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#3
I thing i understand what u did, but the values of the letters are

B = 6
C = 13
D = 5

The at the end, C+D is smaller than 2(B+D)

so here is the new info, Is the probability only a ratio of proportions ?

P(Br+Ye>Bl) = (22-18)/(22-11) = 4/11

Is that it ?
 
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#4
Engage your brain, please. All you need to do is to shift point C in the diagram to the left so that it meets the necessary requirement but that won’t change the diagram’s generality. I’ve revised it accordingly—and note that the shift obviously also affects the point labelled “C+D”.

Now, think about the following aspects:
1: Probability that Br+Ye is in R1;
2: Probability that Br+Ye is in R2;
3: Probability that Br+Ye is in R3;
4: In the case where Br+Ye is in R2, the probability that Br+Ye > Bl, knowing that it’s a uniform distribution; and
5: Which of the above scenarios satisfy Br+Ye > Bl.

Once you have addressed the above, the answer should be obvious. (Hint: it has two components, one of which involves a conditional probability.)
 
#5
Oh i think i got it:

A:third section (R3)
B:second section (R2)
C:Bl<Br+Ye

P(A)=(2(B+D)-(C+D))/2 = 4/11
P(C and B)=P(C|B)*P(B) = (1/4)*(5/11) = 5/44

P(C)=P(A) + P(C and B) = 16/44+5/44 = 21/44
 
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#6
Wrong. With reference to the amended diagram, it should be blindingly obvious that the only time Br+Ye can be less than Bl is when Br+Ye falls in R1 or R2. If Br+Ye is in R1, then Br+Ye < Bl is always true for the given values of B, C and D. If Br+Ye is in R2, how often on average do you expect Br+Ye < Bl, given that every value in that range occurs with equal probability for both Br+Ye and Bl?

Stop. Think. And answer these questions using the symbols B, C and D (not numbers):
(1) What is the probability that Br+Ye is in R1?
(2) What is the probability that Br+Ye is in R2?
(3) Given that Br+Ye is in R2, what is the probability that Br+Ye < Bl?
(4) How must you combine the above probabilities?

Note that if you don’t attempt to answer all of the above questions, you won’t get any further help from me.
 
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