Mixing
Find the probability of $a>b+c$, where $a$, $b$, $c$ are $U(0,1)$
$begingroup$
What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $mathbb{R}$.
probability uniform-distribution
edited Oct 14 '13 at 10:58
martini
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
$endgroup$
add a comment |
$begingroup$
What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $mathbb{R}$.
probability uniform-distribution
edited Oct 14 '13 at 10:58
martini
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
$endgroup$
2
$begingroup$
Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
$endgroup$
– Did
Oct 14 '13 at 9:18
1
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23
add a comment |
$begingroup$
What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $mathbb{R}$.
probability uniform-distribution
edited Oct 14 '13 at 10:58
martini
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
$endgroup$
What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $mathbb{R}$.
probability uniform-distribution
probability uniform-distribution
edited Oct 14 '13 at 10:58
martini
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
edited Oct 14 '13 at 10:58
martini
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
edited Oct 14 '13 at 10:58
martini
70.7k45991
edited Oct 14 '13 at 10:58
martini
70.7k45991
edited Oct 14 '13 at 10:58
martini
70.7k45991
70.7k45991
asked Oct 14 '13 at 9:15
TalitaTalita
22639
asked Oct 14 '13 at 9:15
TalitaTalita
22639
asked Oct 14 '13 at 9:15
TalitaTalita
22639
22639
2
$begingroup$
Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
$endgroup$
– Did
Oct 14 '13 at 9:18
1
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23
add a comment |
2
$begingroup$
Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
$endgroup$
– Did
Oct 14 '13 at 9:18
1
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23
2
2
$begingroup$
Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
$endgroup$
– Did
Oct 14 '13 at 9:18
$begingroup$
Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need.
$endgroup$
– Did
Oct 14 '13 at 9:18
1
1
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Probability as the volume of a pyramid $V = frac{1}{3}Sh = frac{1}{3}cdotfrac{1}{2}cdot1 = frac{1}{6}.$
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
$endgroup$
add a comment |
$begingroup$
Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=frac{1}{n!}sum_{k=0}^{lfloor xrfloor}(-1)^kbinom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
$F'_d(x)=0$ for $xin(-infty,0)cup[2, infty)$. Thus, integral is non-zero only between two intervals:
$x in [0,1). lfloor xrfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{0}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2$$
$$F_d'(x) = x$$
$x in [1,2). lfloor xrfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{1}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2 - frac{1}{2}cdot2(x-1)^2 =-frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - int_{0}^{1}F_a(x), F'_d(x),dx - int_{1}^{2}F_a(x), F'_d(x),dx =$$
$$1 - int_{0}^{1}x, x,dx - int_{1}^{2}1cdot(-x+2),dx = 1 - frac{1}{3} - frac{1}{2} = frac{1}{6}.$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
$endgroup$
add a comment |
$begingroup$
Let's pose that the bounded interval of $mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$int_{max(b+c, 1)}^1 f_a(a) da = 1 - max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$int_{0}^1int_{0}^1 left( 1 - max(b+c, 1) right) f_b(b)f_c(c) db ~dc = 1 - int_{0}^1int_{0}^1max(b+c, 1)db~dc$$
Let's consider $max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$int_{0}^1int_{0}^1max(b+c, 1)db~dc = int_{0}^1left[int_{0}^{1-c}(b+c)db + int_{1-c}^1 1 cdot dbright]dc = $$
$$=int_{0}^1left[frac{(1-c)^2}{2} + c(1-c) + cright]dc = frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - frac{5}{6} = frac{1}{6}$$
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
$endgroup$
add a comment |
$begingroup$
Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:
Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
edited Jan 23 at 20:24
Glorfindel
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
$endgroup$
5
$begingroup$
You might want to continue to mentionI should add that I am one of the authors of mathStatica(or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).
$endgroup$
– Did
Oct 14 '13 at 12:24
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Wouldn't that be boasting?
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– wolfies
Oct 14 '13 at 13:05
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Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
|
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4 Answers
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4 Answers
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$begingroup$
Probability as the volume of a pyramid $V = frac{1}{3}Sh = frac{1}{3}cdotfrac{1}{2}cdot1 = frac{1}{6}.$
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
$endgroup$
add a comment |
$begingroup$
Probability as the volume of a pyramid $V = frac{1}{3}Sh = frac{1}{3}cdotfrac{1}{2}cdot1 = frac{1}{6}.$
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
$endgroup$
add a comment |
$begingroup$
Probability as the volume of a pyramid $V = frac{1}{3}Sh = frac{1}{3}cdotfrac{1}{2}cdot1 = frac{1}{6}.$
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
$endgroup$
Probability as the volume of a pyramid $V = frac{1}{3}Sh = frac{1}{3}cdotfrac{1}{2}cdot1 = frac{1}{6}.$
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
edited Oct 14 '13 at 11:54
Adam Stelmaszczyk
506425
506425
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
answered Oct 14 '13 at 10:52
Kamil CzerskiKamil Czerski
20612
20612
add a comment |
add a comment |
$begingroup$
Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=frac{1}{n!}sum_{k=0}^{lfloor xrfloor}(-1)^kbinom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
$F'_d(x)=0$ for $xin(-infty,0)cup[2, infty)$. Thus, integral is non-zero only between two intervals:
$x in [0,1). lfloor xrfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{0}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2$$
$$F_d'(x) = x$$
$x in [1,2). lfloor xrfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{1}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2 - frac{1}{2}cdot2(x-1)^2 =-frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - int_{0}^{1}F_a(x), F'_d(x),dx - int_{1}^{2}F_a(x), F'_d(x),dx =$$
$$1 - int_{0}^{1}x, x,dx - int_{1}^{2}1cdot(-x+2),dx = 1 - frac{1}{3} - frac{1}{2} = frac{1}{6}.$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
$endgroup$
add a comment |
$begingroup$
Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=frac{1}{n!}sum_{k=0}^{lfloor xrfloor}(-1)^kbinom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
$F'_d(x)=0$ for $xin(-infty,0)cup[2, infty)$. Thus, integral is non-zero only between two intervals:
$x in [0,1). lfloor xrfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{0}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2$$
$$F_d'(x) = x$$
$x in [1,2). lfloor xrfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{1}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2 - frac{1}{2}cdot2(x-1)^2 =-frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - int_{0}^{1}F_a(x), F'_d(x),dx - int_{1}^{2}F_a(x), F'_d(x),dx =$$
$$1 - int_{0}^{1}x, x,dx - int_{1}^{2}1cdot(-x+2),dx = 1 - frac{1}{3} - frac{1}{2} = frac{1}{6}.$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
$endgroup$
add a comment |
$begingroup$
Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=frac{1}{n!}sum_{k=0}^{lfloor xrfloor}(-1)^kbinom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
$F'_d(x)=0$ for $xin(-infty,0)cup[2, infty)$. Thus, integral is non-zero only between two intervals:
$x in [0,1). lfloor xrfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{0}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2$$
$$F_d'(x) = x$$
$x in [1,2). lfloor xrfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{1}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2 - frac{1}{2}cdot2(x-1)^2 =-frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - int_{0}^{1}F_a(x), F'_d(x),dx - int_{1}^{2}F_a(x), F'_d(x),dx =$$
$$1 - int_{0}^{1}x, x,dx - int_{1}^{2}1cdot(-x+2),dx = 1 - frac{1}{3} - frac{1}{2} = frac{1}{6}.$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
$endgroup$
Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=frac{1}{n!}sum_{k=0}^{lfloor xrfloor}(-1)^kbinom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous random variables with cumulative distributions $F_a(x)$ and $F_d(x)$.
So, we can write:
$$P(a > b + c) = P(a > d) = 1 - P(a < d) = 1 - int_{-infty}^{infty} F_a(x), F'_d(x),dx.$$
$F'_d(x)=0$ for $xin(-infty,0)cup[2, infty)$. Thus, integral is non-zero only between two intervals:
$x in [0,1). lfloor xrfloor = 0.$
$$F_a(x) = x$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{0}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2$$
$$F_d'(x) = x$$
$x in [1,2). lfloor xrfloor = 1.$
$$F_a(x) = 1$$
$$F_d(x) = frac{1}{2}sum_{k=0}^{1}(-1)^kbinom{2}{k}(x-k)^2 = frac{1}{2}x^2 - frac{1}{2}cdot2(x-1)^2 =-frac{1}{2}x^2+2x-1$$
$$F_d'(x) = -x+2$$
Therefore:
$$P(a > b + c) = 1 - int_{0}^{1}F_a(x), F'_d(x),dx - int_{1}^{2}F_a(x), F'_d(x),dx =$$
$$1 - int_{0}^{1}x, x,dx - int_{1}^{2}1cdot(-x+2),dx = 1 - frac{1}{3} - frac{1}{2} = frac{1}{6}.$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
answered Oct 14 '13 at 11:40
Adam StelmaszczykAdam Stelmaszczyk
506425
506425
add a comment |
add a comment |
$begingroup$
Let's pose that the bounded interval of $mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$int_{max(b+c, 1)}^1 f_a(a) da = 1 - max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$int_{0}^1int_{0}^1 left( 1 - max(b+c, 1) right) f_b(b)f_c(c) db ~dc = 1 - int_{0}^1int_{0}^1max(b+c, 1)db~dc$$
Let's consider $max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$int_{0}^1int_{0}^1max(b+c, 1)db~dc = int_{0}^1left[int_{0}^{1-c}(b+c)db + int_{1-c}^1 1 cdot dbright]dc = $$
$$=int_{0}^1left[frac{(1-c)^2}{2} + c(1-c) + cright]dc = frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - frac{5}{6} = frac{1}{6}$$
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
$endgroup$
add a comment |
$begingroup$
Let's pose that the bounded interval of $mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$int_{max(b+c, 1)}^1 f_a(a) da = 1 - max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$int_{0}^1int_{0}^1 left( 1 - max(b+c, 1) right) f_b(b)f_c(c) db ~dc = 1 - int_{0}^1int_{0}^1max(b+c, 1)db~dc$$
Let's consider $max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$int_{0}^1int_{0}^1max(b+c, 1)db~dc = int_{0}^1left[int_{0}^{1-c}(b+c)db + int_{1-c}^1 1 cdot dbright]dc = $$
$$=int_{0}^1left[frac{(1-c)^2}{2} + c(1-c) + cright]dc = frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - frac{5}{6} = frac{1}{6}$$
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
$endgroup$
add a comment |
$begingroup$
Let's pose that the bounded interval of $mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$int_{max(b+c, 1)}^1 f_a(a) da = 1 - max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$int_{0}^1int_{0}^1 left( 1 - max(b+c, 1) right) f_b(b)f_c(c) db ~dc = 1 - int_{0}^1int_{0}^1max(b+c, 1)db~dc$$
Let's consider $max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$int_{0}^1int_{0}^1max(b+c, 1)db~dc = int_{0}^1left[int_{0}^{1-c}(b+c)db + int_{1-c}^1 1 cdot dbright]dc = $$
$$=int_{0}^1left[frac{(1-c)^2}{2} + c(1-c) + cright]dc = frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - frac{5}{6} = frac{1}{6}$$
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
$endgroup$
Let's pose that the bounded interval of $mathbb{R}$ is $I = [0, 1]$.
Fixed $b$ and $c$, we have to find out the conditional probability that $a > b+c$. Namely:
$$int_{max(b+c, 1)}^1 f_a(a) da = 1 - max(b+c, 1)$$
Then, we have to integrate with respect to $b$ and $c$:
$$int_{0}^1int_{0}^1 left( 1 - max(b+c, 1) right) f_b(b)f_c(c) db ~dc = 1 - int_{0}^1int_{0}^1max(b+c, 1)db~dc$$
Let's consider $max(b+c, 1)$. It is equal to $b+c$ when $b < 1 - c$, and $1$ elsewhere. Then:
$$int_{0}^1int_{0}^1max(b+c, 1)db~dc = int_{0}^1left[int_{0}^{1-c}(b+c)db + int_{1-c}^1 1 cdot dbright]dc = $$
$$=int_{0}^1left[frac{(1-c)^2}{2} + c(1-c) + cright]dc = frac{5}{6}$$
Finally, the probabilty you are looking for is
$$1 - frac{5}{6} = frac{1}{6}$$
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
edited Oct 14 '13 at 11:00
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
answered Oct 14 '13 at 9:30
the_candymanthe_candyman
8,97832145
8,97832145
add a comment |
add a comment |
$begingroup$
Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:
Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
edited Jan 23 at 20:24
Glorfindel
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
$endgroup$
5
$begingroup$
You might want to continue to mentionI should add that I am one of the authors of mathStatica(or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).
$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
|
show 1 more comment
$begingroup$
Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:
Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
edited Jan 23 at 20:24
Glorfindel
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
$endgroup$
5
$begingroup$
You might want to continue to mentionI should add that I am one of the authors of mathStatica(or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).
$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
|
show 1 more comment
$begingroup$
Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:
Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
edited Jan 23 at 20:24
Glorfindel
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
$endgroup$
Others have provided graphical, manual integration and distribution theory approaches ---> so here's an automated computer algebra system approach, as a one-liner:
Easy and simple ... calculated here using the mathStatica add-on to Mathematica (I am one of the authors of the former); presumably Maple or other computer algebra packages could handle it too (I am not an author of those packages).
edited Jan 23 at 20:24
Glorfindel
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
edited Jan 23 at 20:24
Glorfindel
3,42981830
edited Jan 23 at 20:24
Glorfindel
3,42981830
edited Jan 23 at 20:24
Glorfindel
3,42981830
3,42981830
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
answered Oct 14 '13 at 11:59
wolfieswolfies
4,2392923
4,2392923
5
$begingroup$
You might want to continue to mentionI should add that I am one of the authors of mathStatica(or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).
$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
|
show 1 more comment
5
$begingroup$
You might want to continue to mentionI should add that I am one of the authors of mathStatica(or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).
$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
5
5
$begingroup$
You might want to continue to mention
I should add that I am one of the authors of mathStatica (or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
You might want to continue to mention
I should add that I am one of the authors of mathStatica (or a similar formulation) when posting answers based on this software (in three recent answers, you forgot to do so).$endgroup$
– Did
Oct 14 '13 at 12:24
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Wouldn't that be boasting?
$endgroup$
– wolfies
Oct 14 '13 at 13:05
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
$begingroup$
Are you going to add the mention to this answer and to the others, or not?
$endgroup$
– Did
Oct 14 '13 at 17:03
2
2
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@wolfies: it doesn't seem like boasting. There is a difference between an impartial endorsement and a paid endorsement. Granted, this is somewhere in-between, but if you don't say something, it looks like an impartial endorsement, which it really is not.
$endgroup$
– robjohn♦
Oct 14 '13 at 21:16
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
$begingroup$
@robjohn No problem.
$endgroup$
– wolfies
Oct 15 '13 at 3:55
|
show 1 more comment
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– Did
Oct 14 '13 at 9:18
1
$begingroup$
what interval is involved here? It is important to know that. For instance if it is something like $left(5,6right)$ then automatically $a<b+c$. So the probability you mention is $0$ in that case.
$endgroup$
– drhab
Oct 14 '13 at 9:23