Mixing
We break a unit length stick at two randomly chosen points
$begingroup$
We break a unit length stick at two randomly chosen points.
(a) Describe the natural probability space for this experiment.
(b) Consider the event that one of the resulting three sticks is larger than $frac{1}{2}.$ Give a description of this event in the sample space of your probability space.
(c) Find the probability of the event in (b).
First we note that the stick can be model by the unit interval $[0,1].$
For part (a), I constructed the following probability space
$$Omega={ omega=[0,x_1]cup[x_1,x_2]cup[x_2,1] : 0leq x_{i}<x_{i+1}leq 1 space text{for} space i=1}.$$
For part (b), we call our event $A$. Then we have that
$$A={ [0,x_1]cup[x_1,x_2]cup[x_2,1]: |x_1-0|>frac{1}{2} space text{or} space |x_1-x_2|>frac{1}{2} space text{or} space |1-x_2|>frac{1}{2} }.$$
For part (c), we can consider the complement, which is the event where all three resulting sticks have length smaller than $frac{1}{2}.$
I'm having difficulty conceptualizing this case. I think we it is easier to deal the complement, and then use the axioms, but I do not see how we could write the complement as a union of disjoint sets.
Any help is appreciated.
Is the approach in (a) and (b) above correct?
probability probability-theory
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
add a comment |
$begingroup$
We break a unit length stick at two randomly chosen points.
(a) Describe the natural probability space for this experiment.
(b) Consider the event that one of the resulting three sticks is larger than $frac{1}{2}.$ Give a description of this event in the sample space of your probability space.
(c) Find the probability of the event in (b).
First we note that the stick can be model by the unit interval $[0,1].$
For part (a), I constructed the following probability space
$$Omega={ omega=[0,x_1]cup[x_1,x_2]cup[x_2,1] : 0leq x_{i}<x_{i+1}leq 1 space text{for} space i=1}.$$
For part (b), we call our event $A$. Then we have that
$$A={ [0,x_1]cup[x_1,x_2]cup[x_2,1]: |x_1-0|>frac{1}{2} space text{or} space |x_1-x_2|>frac{1}{2} space text{or} space |1-x_2|>frac{1}{2} }.$$
For part (c), we can consider the complement, which is the event where all three resulting sticks have length smaller than $frac{1}{2}.$
I'm having difficulty conceptualizing this case. I think we it is easier to deal the complement, and then use the axioms, but I do not see how we could write the complement as a union of disjoint sets.
Any help is appreciated.
Is the approach in (a) and (b) above correct?
probability probability-theory
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35
add a comment |
$begingroup$
We break a unit length stick at two randomly chosen points.
(a) Describe the natural probability space for this experiment.
(b) Consider the event that one of the resulting three sticks is larger than $frac{1}{2}.$ Give a description of this event in the sample space of your probability space.
(c) Find the probability of the event in (b).
First we note that the stick can be model by the unit interval $[0,1].$
For part (a), I constructed the following probability space
$$Omega={ omega=[0,x_1]cup[x_1,x_2]cup[x_2,1] : 0leq x_{i}<x_{i+1}leq 1 space text{for} space i=1}.$$
For part (b), we call our event $A$. Then we have that
$$A={ [0,x_1]cup[x_1,x_2]cup[x_2,1]: |x_1-0|>frac{1}{2} space text{or} space |x_1-x_2|>frac{1}{2} space text{or} space |1-x_2|>frac{1}{2} }.$$
For part (c), we can consider the complement, which is the event where all three resulting sticks have length smaller than $frac{1}{2}.$
I'm having difficulty conceptualizing this case. I think we it is easier to deal the complement, and then use the axioms, but I do not see how we could write the complement as a union of disjoint sets.
Any help is appreciated.
Is the approach in (a) and (b) above correct?
probability probability-theory
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
$endgroup$
We break a unit length stick at two randomly chosen points.
(a) Describe the natural probability space for this experiment.
(b) Consider the event that one of the resulting three sticks is larger than $frac{1}{2}.$ Give a description of this event in the sample space of your probability space.
(c) Find the probability of the event in (b).
First we note that the stick can be model by the unit interval $[0,1].$
For part (a), I constructed the following probability space
$$Omega={ omega=[0,x_1]cup[x_1,x_2]cup[x_2,1] : 0leq x_{i}<x_{i+1}leq 1 space text{for} space i=1}.$$
For part (b), we call our event $A$. Then we have that
$$A={ [0,x_1]cup[x_1,x_2]cup[x_2,1]: |x_1-0|>frac{1}{2} space text{or} space |x_1-x_2|>frac{1}{2} space text{or} space |1-x_2|>frac{1}{2} }.$$
For part (c), we can consider the complement, which is the event where all three resulting sticks have length smaller than $frac{1}{2}.$
I'm having difficulty conceptualizing this case. I think we it is easier to deal the complement, and then use the axioms, but I do not see how we could write the complement as a union of disjoint sets.
Any help is appreciated.
Is the approach in (a) and (b) above correct?
probability probability-theory
probability probability-theory
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
edited Feb 1 at 7:28
Gaby Alfonso
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
asked Feb 1 at 7:06
Gaby AlfonsoGaby Alfonso
1,1951418
1,1951418
$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35
add a comment |
$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35
$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35
add a comment |
1 Answer
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$begingroup$
(a)
$$
(Omega,mathcal{F},mathsf{P})=([0,1]^2,mathcal{B}([0,1]^2),lambdaotimeslambda),
$$
where $lambda$ is the Lebesgue measure on $[0,1]$.
(b) Letting $omega_{(1)}=omega_1wedgeomega_2$ and $omega_{(2)}=omega_1veeomega_2$, the required event is
$$
A=A_1cup A_2cup A_3,
$$
where
begin{align}
A_1&={omegainOmega:omega_{(1)}>1/2},\
A_2&={omegainOmega:omega_{(2)}<1/2}, quadtext{and} \
A_3&={omegainOmega:omega_{(2)}-omega_{(1)}>1/2}.
end{align}
(c) Since $A_1$, $A_2$, and $A_3$ are disjoint,
begin{align}
mathsf{P}(A)&=mathsf{P}(A_1)+mathsf{P}(A_2)+mathsf{P}(A_3) \
&=(1/2)^2+(1/2)^2+(1/2)^2=3/4.
end{align}
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
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1 Answer
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$begingroup$
(a)
$$
(Omega,mathcal{F},mathsf{P})=([0,1]^2,mathcal{B}([0,1]^2),lambdaotimeslambda),
$$
where $lambda$ is the Lebesgue measure on $[0,1]$.
(b) Letting $omega_{(1)}=omega_1wedgeomega_2$ and $omega_{(2)}=omega_1veeomega_2$, the required event is
$$
A=A_1cup A_2cup A_3,
$$
where
begin{align}
A_1&={omegainOmega:omega_{(1)}>1/2},\
A_2&={omegainOmega:omega_{(2)}<1/2}, quadtext{and} \
A_3&={omegainOmega:omega_{(2)}-omega_{(1)}>1/2}.
end{align}
(c) Since $A_1$, $A_2$, and $A_3$ are disjoint,
begin{align}
mathsf{P}(A)&=mathsf{P}(A_1)+mathsf{P}(A_2)+mathsf{P}(A_3) \
&=(1/2)^2+(1/2)^2+(1/2)^2=3/4.
end{align}
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
$begingroup$
(a)
$$
(Omega,mathcal{F},mathsf{P})=([0,1]^2,mathcal{B}([0,1]^2),lambdaotimeslambda),
$$
where $lambda$ is the Lebesgue measure on $[0,1]$.
(b) Letting $omega_{(1)}=omega_1wedgeomega_2$ and $omega_{(2)}=omega_1veeomega_2$, the required event is
$$
A=A_1cup A_2cup A_3,
$$
where
begin{align}
A_1&={omegainOmega:omega_{(1)}>1/2},\
A_2&={omegainOmega:omega_{(2)}<1/2}, quadtext{and} \
A_3&={omegainOmega:omega_{(2)}-omega_{(1)}>1/2}.
end{align}
(c) Since $A_1$, $A_2$, and $A_3$ are disjoint,
begin{align}
mathsf{P}(A)&=mathsf{P}(A_1)+mathsf{P}(A_2)+mathsf{P}(A_3) \
&=(1/2)^2+(1/2)^2+(1/2)^2=3/4.
end{align}
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
$begingroup$
(a)
$$
(Omega,mathcal{F},mathsf{P})=([0,1]^2,mathcal{B}([0,1]^2),lambdaotimeslambda),
$$
where $lambda$ is the Lebesgue measure on $[0,1]$.
(b) Letting $omega_{(1)}=omega_1wedgeomega_2$ and $omega_{(2)}=omega_1veeomega_2$, the required event is
$$
A=A_1cup A_2cup A_3,
$$
where
begin{align}
A_1&={omegainOmega:omega_{(1)}>1/2},\
A_2&={omegainOmega:omega_{(2)}<1/2}, quadtext{and} \
A_3&={omegainOmega:omega_{(2)}-omega_{(1)}>1/2}.
end{align}
(c) Since $A_1$, $A_2$, and $A_3$ are disjoint,
begin{align}
mathsf{P}(A)&=mathsf{P}(A_1)+mathsf{P}(A_2)+mathsf{P}(A_3) \
&=(1/2)^2+(1/2)^2+(1/2)^2=3/4.
end{align}
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
$endgroup$
(a)
$$
(Omega,mathcal{F},mathsf{P})=([0,1]^2,mathcal{B}([0,1]^2),lambdaotimeslambda),
$$
where $lambda$ is the Lebesgue measure on $[0,1]$.
(b) Letting $omega_{(1)}=omega_1wedgeomega_2$ and $omega_{(2)}=omega_1veeomega_2$, the required event is
$$
A=A_1cup A_2cup A_3,
$$
where
begin{align}
A_1&={omegainOmega:omega_{(1)}>1/2},\
A_2&={omegainOmega:omega_{(2)}<1/2}, quadtext{and} \
A_3&={omegainOmega:omega_{(2)}-omega_{(1)}>1/2}.
end{align}
(c) Since $A_1$, $A_2$, and $A_3$ are disjoint,
begin{align}
mathsf{P}(A)&=mathsf{P}(A_1)+mathsf{P}(A_2)+mathsf{P}(A_3) \
&=(1/2)^2+(1/2)^2+(1/2)^2=3/4.
end{align}
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
edited Feb 1 at 10:16
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
answered Feb 1 at 10:08
d.k.o.d.k.o.
10.6k730
10.6k730
add a comment |
add a comment |
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$begingroup$
I would probably model the two cuts as a uniformly chosen point in $[0,1]times[0,1]$.
$endgroup$
– Arthur
Feb 1 at 7:28
$begingroup$
I would model it with X~U[0,1] and you have samples $X_1$ and $X_2$. Than b is calculating $P(X_1<frac{1}{2} wedge X_2 <frac{1}{2}) + P(X_1>frac{1}{2} wedge X_2 >frac{1}{2}) + P(|X_1-X_2|>frac{1}{2})$ Note you can sum them because 2 events cannot happen together
$endgroup$
– Shaq
Feb 1 at 7:35