Mixing
Probability distribution and floor functions: is this the right approach?
$begingroup$
The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.
(a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?
(b) Find the limits of these probabilities as $n rightarrowinfty$.
(c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?
MY ATTEMPT
The book gives the following answers to each case:
(a) It uses the floor function in order to solve this problem. Precisely, we have:
begin{align*}
textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
From whence we obtain that
begin{align*}
textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
(b) I do not know how to handle limits with floor function. But it says that it should be
begin{align*}
lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
end{align*}
(c) It is the same answer as given at (a).
Is there another approach to this problem? Thanks in advance.
probability probability-theory proof-verification
asked Jan 11 at 19:48
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.
(a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?
(b) Find the limits of these probabilities as $n rightarrowinfty$.
(c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?
MY ATTEMPT
The book gives the following answers to each case:
(a) It uses the floor function in order to solve this problem. Precisely, we have:
begin{align*}
textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
From whence we obtain that
begin{align*}
textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
(b) I do not know how to handle limits with floor function. But it says that it should be
begin{align*}
lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
end{align*}
(c) It is the same answer as given at (a).
Is there another approach to this problem? Thanks in advance.
probability probability-theory proof-verification
asked Jan 11 at 19:48
user1337user1337
46110
$endgroup$
add a comment |
$begingroup$
The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.
(a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?
(b) Find the limits of these probabilities as $n rightarrowinfty$.
(c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?
MY ATTEMPT
The book gives the following answers to each case:
(a) It uses the floor function in order to solve this problem. Precisely, we have:
begin{align*}
textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
From whence we obtain that
begin{align*}
textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
(b) I do not know how to handle limits with floor function. But it says that it should be
begin{align*}
lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
end{align*}
(c) It is the same answer as given at (a).
Is there another approach to this problem? Thanks in advance.
probability probability-theory proof-verification
asked Jan 11 at 19:48
user1337user1337
46110
$endgroup$
The consecutive integers $1, 2,ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.
(a) What are $textbf{P}(D_{3})$, $textbf{P}(D_{4})$, $textbf{P}(D_{3}cup D_{4})$ and $textbf{P}(D_{3}cap D_{4})$?
(b) Find the limits of these probabilities as $n rightarrowinfty$.
(c) What would your answers be if the $n$ consecutive numbers began at a number $a neq 1$?
MY ATTEMPT
The book gives the following answers to each case:
(a) It uses the floor function in order to solve this problem. Precisely, we have:
begin{align*}
textbf{P}(D_{3}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor}{n},quad textbf{P}(D_{4}) = frac{displaystyleleftlfloorfrac{n}{4} rightrfloor}{n},quad textbf{P}(D_{3}cap D_{4}) = frac{displaystyleleftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
From whence we obtain that
begin{align*}
textbf{P}(D_{3}cup D_{4}) = frac{displaystyleleftlfloorfrac{n}{3} rightrfloor + leftlfloorfrac{n}{4}rightrfloor - leftlfloorfrac{n}{12} rightrfloor}{n}
end{align*}
(b) I do not know how to handle limits with floor function. But it says that it should be
begin{align*}
lim_{nrightarrow+infty}textbf{P}(D_{3}) = frac{1}{3}, quad lim_{nrightarrow+infty}textbf{P}(D_{4}) = frac{1}{4}, quadlim_{nrightarrow+infty}textbf{P}(D_{3}cap D_{4}) = frac{1}{12}
end{align*}
(c) It is the same answer as given at (a).
Is there another approach to this problem? Thanks in advance.
probability probability-theory proof-verification
probability probability-theory proof-verification
asked Jan 11 at 19:48
user1337user1337
46110
asked Jan 11 at 19:48
user1337user1337
46110
edited Jan 14 at 2:26
user1337
asked Jan 11 at 19:48
user1337user1337
46110
asked Jan 11 at 19:48
user1337user1337
46110
asked Jan 11 at 19:48
user1337user1337
46110
46110
add a comment |
add a comment |
1 Answer
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$begingroup$
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
and if $n=3k+2$
for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
$$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.
answered Jan 11 at 20:04
pwerthpwerth
3,233417
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add a comment |
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1 Answer
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$begingroup$
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
and if $n=3k+2$
for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
$$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.
answered Jan 11 at 20:04
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
and if $n=3k+2$
for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
$$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.
answered Jan 11 at 20:04
pwerthpwerth
3,233417
$endgroup$
add a comment |
$begingroup$
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
and if $n=3k+2$
for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
$$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.
answered Jan 11 at 20:04
pwerthpwerth
3,233417
$endgroup$
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+1}$$
and if $n=3k+2$
for some $k$, then
$$P(D_{3})=frac{k}{n}=frac{k}{3k+2}$$
Notice that the limit of each of these as $ntoinfty$ is clearly $1/3$.
Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is
$$frac{1}{3}(frac{1}{3}+frac{1}{3}+frac{1}{3})=frac{1}{3}$$
as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.
answered Jan 11 at 20:04
pwerthpwerth
3,233417
answered Jan 11 at 20:04
pwerthpwerth
3,233417
answered Jan 11 at 20:04
pwerthpwerth
3,233417
answered Jan 11 at 20:04
pwerthpwerth
3,233417
3,233417
add a comment |
add a comment |
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