$n $ balls and $n$ bins [closed]
$begingroup$
Suppose that $n$ balls are thrown independently and uniformly at random
into $n$ bins.
EDIT: Sorry, i write wrong question
Correct is: Find the conditional probability that bin $i$ has one ball given that exactly one ball fell into the first three bins.
probability
asked Jan 13 at 19:34
user3681666user3681666
133
$endgroup$
closed as off-topic by Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho Jan 14 at 6:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose that $n$ balls are thrown independently and uniformly at random
into $n$ bins.
EDIT: Sorry, i write wrong question
Correct is: Find the conditional probability that bin $i$ has one ball given that exactly one ball fell into the first three bins.
probability
asked Jan 13 at 19:34
user3681666user3681666
133
$endgroup$
closed as off-topic by Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho Jan 14 at 6:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose that $n$ balls are thrown independently and uniformly at random
into $n$ bins.
EDIT: Sorry, i write wrong question
Correct is: Find the conditional probability that bin $i$ has one ball given that exactly one ball fell into the first three bins.
probability
asked Jan 13 at 19:34
user3681666user3681666
133
$endgroup$
Suppose that $n$ balls are thrown independently and uniformly at random
into $n$ bins.
EDIT: Sorry, i write wrong question
Correct is: Find the conditional probability that bin $i$ has one ball given that exactly one ball fell into the first three bins.
probability
probability
asked Jan 13 at 19:34
user3681666user3681666
133
asked Jan 13 at 19:34
user3681666user3681666
133
edited Jan 14 at 11:32
user3681666
asked Jan 13 at 19:34
user3681666user3681666
133
asked Jan 13 at 19:34
user3681666user3681666
133
asked Jan 13 at 19:34
user3681666user3681666
133
133
closed as off-topic by Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho Jan 14 at 6:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho Jan 14 at 6:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, max_zorn, Leucippus, Eevee Trainer, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One approach: let $X_1$ and $X_2$ be the number of balls in bins 1 and 2 respectively. Then
$$P(X_1 < X_2) + P(X_1 > X_2) + P(X_1 = X_2) = 1.$$
By symmetry, $P(X_1 < X_2) = P(X_1 > X_2)$, so we have
$$P(X_1 > X_2) = frac{1}{2} (1 - P(X_1 = X_2)).$$
Now compute $P(X_1 = X_2)$.
It may help to write it as $$P(X_1 = X_2) = sum_{k=0}^{lfloor n/2 rfloor} P(X_1 = X_2 = k).$$
begin{align}P(X_1 = X_2 = k) &= P(text{$k$ balls in bin 1, $k$ balls in bin 2, $n-2k$ balls in other bins})\ &= left(binom{n}{k}frac{1}{n^k}right)^2 left(frac{n-2}{n}right)^{n-2k}end{align}
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One approach: let $X_1$ and $X_2$ be the number of balls in bins 1 and 2 respectively. Then
$$P(X_1 < X_2) + P(X_1 > X_2) + P(X_1 = X_2) = 1.$$
By symmetry, $P(X_1 < X_2) = P(X_1 > X_2)$, so we have
$$P(X_1 > X_2) = frac{1}{2} (1 - P(X_1 = X_2)).$$
Now compute $P(X_1 = X_2)$.
It may help to write it as $$P(X_1 = X_2) = sum_{k=0}^{lfloor n/2 rfloor} P(X_1 = X_2 = k).$$
begin{align}P(X_1 = X_2 = k) &= P(text{$k$ balls in bin 1, $k$ balls in bin 2, $n-2k$ balls in other bins})\ &= left(binom{n}{k}frac{1}{n^k}right)^2 left(frac{n-2}{n}right)^{n-2k}end{align}
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
add a comment |
$begingroup$
One approach: let $X_1$ and $X_2$ be the number of balls in bins 1 and 2 respectively. Then
$$P(X_1 < X_2) + P(X_1 > X_2) + P(X_1 = X_2) = 1.$$
By symmetry, $P(X_1 < X_2) = P(X_1 > X_2)$, so we have
$$P(X_1 > X_2) = frac{1}{2} (1 - P(X_1 = X_2)).$$
Now compute $P(X_1 = X_2)$.
It may help to write it as $$P(X_1 = X_2) = sum_{k=0}^{lfloor n/2 rfloor} P(X_1 = X_2 = k).$$
begin{align}P(X_1 = X_2 = k) &= P(text{$k$ balls in bin 1, $k$ balls in bin 2, $n-2k$ balls in other bins})\ &= left(binom{n}{k}frac{1}{n^k}right)^2 left(frac{n-2}{n}right)^{n-2k}end{align}
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
add a comment |
$begingroup$
One approach: let $X_1$ and $X_2$ be the number of balls in bins 1 and 2 respectively. Then
$$P(X_1 < X_2) + P(X_1 > X_2) + P(X_1 = X_2) = 1.$$
By symmetry, $P(X_1 < X_2) = P(X_1 > X_2)$, so we have
$$P(X_1 > X_2) = frac{1}{2} (1 - P(X_1 = X_2)).$$
Now compute $P(X_1 = X_2)$.
It may help to write it as $$P(X_1 = X_2) = sum_{k=0}^{lfloor n/2 rfloor} P(X_1 = X_2 = k).$$
begin{align}P(X_1 = X_2 = k) &= P(text{$k$ balls in bin 1, $k$ balls in bin 2, $n-2k$ balls in other bins})\ &= left(binom{n}{k}frac{1}{n^k}right)^2 left(frac{n-2}{n}right)^{n-2k}end{align}
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
$endgroup$
One approach: let $X_1$ and $X_2$ be the number of balls in bins 1 and 2 respectively. Then
$$P(X_1 < X_2) + P(X_1 > X_2) + P(X_1 = X_2) = 1.$$
By symmetry, $P(X_1 < X_2) = P(X_1 > X_2)$, so we have
$$P(X_1 > X_2) = frac{1}{2} (1 - P(X_1 = X_2)).$$
Now compute $P(X_1 = X_2)$.
It may help to write it as $$P(X_1 = X_2) = sum_{k=0}^{lfloor n/2 rfloor} P(X_1 = X_2 = k).$$
begin{align}P(X_1 = X_2 = k) &= P(text{$k$ balls in bin 1, $k$ balls in bin 2, $n-2k$ balls in other bins})\ &= left(binom{n}{k}frac{1}{n^k}right)^2 left(frac{n-2}{n}right)^{n-2k}end{align}
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
edited Jan 13 at 21:22
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
answered Jan 13 at 20:00
angryavianangryavian
40.6k23380
40.6k23380
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
add a comment |
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
$begingroup$
Thanks for your answer :) But I have no idea how can I solve this sum :/ Can you give some hints?
$endgroup$
– user3681666
Jan 13 at 21:19
add a comment |
