Please explain how to find the probability that it passes through the mesh
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A sieve is made of a square mesh of wires. Each wire has diameter d, and the holes
in the mesh are squares whose side length is w. A spherical particle of radius r is
dropped on the mesh. What is the probability that it fails to pass through if it is dropped n times? (Calculations such as these are relevant to the theory of sieving for analyzing the size distribution of particulate matter.)
my answer is $frac{(w-d-2r)^2}{(w-d)^2}$ am i right?help,i dont't known how to proceed
probability proof-verification
edited Nov 17 at 16:13
klirk
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
add a comment |
up vote
0
down vote
favorite
A sieve is made of a square mesh of wires. Each wire has diameter d, and the holes
in the mesh are squares whose side length is w. A spherical particle of radius r is
dropped on the mesh. What is the probability that it fails to pass through if it is dropped n times? (Calculations such as these are relevant to the theory of sieving for analyzing the size distribution of particulate matter.)
my answer is $frac{(w-d-2r)^2}{(w-d)^2}$ am i right?help,i dont't known how to proceed
probability proof-verification
edited Nov 17 at 16:13
klirk
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A sieve is made of a square mesh of wires. Each wire has diameter d, and the holes
in the mesh are squares whose side length is w. A spherical particle of radius r is
dropped on the mesh. What is the probability that it fails to pass through if it is dropped n times? (Calculations such as these are relevant to the theory of sieving for analyzing the size distribution of particulate matter.)
my answer is $frac{(w-d-2r)^2}{(w-d)^2}$ am i right?help,i dont't known how to proceed
probability proof-verification
edited Nov 17 at 16:13
klirk
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
A sieve is made of a square mesh of wires. Each wire has diameter d, and the holes
in the mesh are squares whose side length is w. A spherical particle of radius r is
dropped on the mesh. What is the probability that it fails to pass through if it is dropped n times? (Calculations such as these are relevant to the theory of sieving for analyzing the size distribution of particulate matter.)
my answer is $frac{(w-d-2r)^2}{(w-d)^2}$ am i right?help,i dont't known how to proceed
probability proof-verification
probability proof-verification
edited Nov 17 at 16:13
klirk
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
edited Nov 17 at 16:13
klirk
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
edited Nov 17 at 16:13
klirk
2,220428
edited Nov 17 at 16:13
klirk
2,220428
edited Nov 17 at 16:13
klirk
2,220428
2,220428
asked Nov 17 at 15:58
liuzhiwei
11
asked Nov 17 at 15:58
liuzhiwei
11
asked Nov 17 at 15:58
liuzhiwei
11
11
I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46
add a comment |
I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46
I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
If $w$ is the width of a mesh hole, then $(frac{1}{2}d+w+frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = frac{A_{pass}}{A_{total}} = frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = left(1-dfrac{(w-2r)^2}{(w+d)^2}right)^n$$
answered Nov 18 at 15:29
Cuspy Code
32626
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $w$ is the width of a mesh hole, then $(frac{1}{2}d+w+frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = frac{A_{pass}}{A_{total}} = frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = left(1-dfrac{(w-2r)^2}{(w+d)^2}right)^n$$
answered Nov 18 at 15:29
Cuspy Code
32626
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
add a comment |
up vote
0
down vote
If $w$ is the width of a mesh hole, then $(frac{1}{2}d+w+frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = frac{A_{pass}}{A_{total}} = frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = left(1-dfrac{(w-2r)^2}{(w+d)^2}right)^n$$
answered Nov 18 at 15:29
Cuspy Code
32626
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
add a comment |
up vote
0
down vote
up vote
0
down vote
If $w$ is the width of a mesh hole, then $(frac{1}{2}d+w+frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = frac{A_{pass}}{A_{total}} = frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = left(1-dfrac{(w-2r)^2}{(w+d)^2}right)^n$$
answered Nov 18 at 15:29
Cuspy Code
32626
If $w$ is the width of a mesh hole, then $(frac{1}{2}d+w+frac{1}{2}d)=(w+d)$ is the distance from one wire midpoint to the midpoint of the wire on the opposite side of the hole. If $N$ is the number of holes, then the total area of the mesh is
$$A_{total} = N, (w+d)^2$$
In order to pass through the hole without touching a wire, the center of the spherical particle needs to be positioned within a smaller square that has the side $(w-2r)$. The total area that allows pass-through is therefore
$$A_{pass} = N, (w-2r)^2$$
For a single drop, the probability of passing through is
$$P_{pass} = frac{A_{pass}}{A_{total}} = frac{(w-2r)^2}{(w+d)^2}$$
And the probability of failing is therefore
$$P_{fail} = 1 - P_{pass}$$
Assuming each drop is an independent event, the probability of failing $n$ drops is
$$(P_{fail})^n = (1-P_{pass})^n = left(1-dfrac{(w-2r)^2}{(w+d)^2}right)^n$$
answered Nov 18 at 15:29
Cuspy Code
32626
edited Nov 20 at 17:47
answered Nov 18 at 15:29
Cuspy Code
32626
answered Nov 18 at 15:29
Cuspy Code
32626
answered Nov 18 at 15:29
Cuspy Code
32626
32626
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
add a comment |
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
Thanks for answering this question, that was cool
– liuzhiwei
Nov 20 at 4:07
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
bring me new idea,I suspect the right $A_{pass} is N(w-2r)^2$
– liuzhiwei
Nov 20 at 4:45
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
@liuzhiwei You are right! My mistake. I was thinking of diameters, just like for the wires. I have now edited my answer to correct this.
– Cuspy Code
Nov 20 at 17:49
add a comment |
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I am not familiar with the formulas needed to compute this, as most users probably are not. If you could post your way of solution, then we can check.
– klirk
Nov 17 at 16:12
Before we can answer this question mathematically, I think it is important to know what happens if the spherical particle narrowly makes contact with the wire. Does it bounce back from the sieve, or does it slip through?
– M. Wind
Nov 17 at 16:54
Is $w$ the width of the hole, or is it the distance from one wire center to the next? Also, the probability for failing to pass through (once) should be proportional to the area of the non-hole part, not the area of the hole.
– Cuspy Code
Nov 17 at 18:00
this question comes from the third chapter of "Mathematical .statistics.and data.analysis,.3rd" joint distributions , exercise. No more detail were given
– liuzhiwei
Nov 18 at 2:31
my answer based on hypothesis that the object is stay where it lands passing through meas passing through without toughing the wires.
– liuzhiwei
Nov 18 at 2:46