Mixing
$ X_i $ is discrete random variable, Compute $ sum X_i = 97 $
$begingroup$
Let $ X_1, X_2, , dots , X_{10} $ be a discrete random variable with uniform distribution between $ 0 $ to $ 10 $. Compute $ P{ sum_{i=1}^{10} X_i = 97 } $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$ P{ sum_{i=1}^{10} X_i = 97 } = {10 choose 1 } cdot frac{1}{11}^{10} + {10 choose 2} cdot frac{1}{11}^{10} + {10 choose 3} cdot frac{1}{11}^{10} $$
probability uniform-distribution
asked Jan 3 at 9:53
bm1125bm1125
64016
$endgroup$
add a comment |
$begingroup$
Let $ X_1, X_2, , dots , X_{10} $ be a discrete random variable with uniform distribution between $ 0 $ to $ 10 $. Compute $ P{ sum_{i=1}^{10} X_i = 97 } $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$ P{ sum_{i=1}^{10} X_i = 97 } = {10 choose 1 } cdot frac{1}{11}^{10} + {10 choose 2} cdot frac{1}{11}^{10} + {10 choose 3} cdot frac{1}{11}^{10} $$
probability uniform-distribution
asked Jan 3 at 9:53
bm1125bm1125
64016
$endgroup$
$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58
add a comment |
$begingroup$
Let $ X_1, X_2, , dots , X_{10} $ be a discrete random variable with uniform distribution between $ 0 $ to $ 10 $. Compute $ P{ sum_{i=1}^{10} X_i = 97 } $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$ P{ sum_{i=1}^{10} X_i = 97 } = {10 choose 1 } cdot frac{1}{11}^{10} + {10 choose 2} cdot frac{1}{11}^{10} + {10 choose 3} cdot frac{1}{11}^{10} $$
probability uniform-distribution
asked Jan 3 at 9:53
bm1125bm1125
64016
$endgroup$
Let $ X_1, X_2, , dots , X_{10} $ be a discrete random variable with uniform distribution between $ 0 $ to $ 10 $. Compute $ P{ sum_{i=1}^{10} X_i = 97 } $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?
$$ P{ sum_{i=1}^{10} X_i = 97 } = {10 choose 1 } cdot frac{1}{11}^{10} + {10 choose 2} cdot frac{1}{11}^{10} + {10 choose 3} cdot frac{1}{11}^{10} $$
probability uniform-distribution
probability uniform-distribution
asked Jan 3 at 9:53
bm1125bm1125
64016
asked Jan 3 at 9:53
bm1125bm1125
64016
asked Jan 3 at 9:53
bm1125bm1125
64016
asked Jan 3 at 9:53
bm1125bm1125
64016
asked Jan 3 at 9:53
bm1125bm1125
64016
64016
$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58
add a comment |
$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58
$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58
add a comment |
1 Answer
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$begingroup$
Your solution is almost correct.
The correct solution is:$$11^{-10}left(frac{10!}{9!1!}+frac{10!}{8!1!1!}+frac{10!}{7!3!}right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $binom{10}{2}=frac{10!}{8!2!}$ must be replaced by $frac{10!}{8!1!1!}$.
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
$endgroup$
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Your solution is almost correct.
The correct solution is:$$11^{-10}left(frac{10!}{9!1!}+frac{10!}{8!1!1!}+frac{10!}{7!3!}right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $binom{10}{2}=frac{10!}{8!2!}$ must be replaced by $frac{10!}{8!1!1!}$.
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
$endgroup$
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
add a comment |
$begingroup$
Your solution is almost correct.
The correct solution is:$$11^{-10}left(frac{10!}{9!1!}+frac{10!}{8!1!1!}+frac{10!}{7!3!}right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $binom{10}{2}=frac{10!}{8!2!}$ must be replaced by $frac{10!}{8!1!1!}$.
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
$endgroup$
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
add a comment |
$begingroup$
Your solution is almost correct.
The correct solution is:$$11^{-10}left(frac{10!}{9!1!}+frac{10!}{8!1!1!}+frac{10!}{7!3!}right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $binom{10}{2}=frac{10!}{8!2!}$ must be replaced by $frac{10!}{8!1!1!}$.
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
$endgroup$
Your solution is almost correct.
The correct solution is:$$11^{-10}left(frac{10!}{9!1!}+frac{10!}{8!1!1!}+frac{10!}{7!3!}right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $binom{10}{2}=frac{10!}{8!2!}$ must be replaced by $frac{10!}{8!1!1!}$.
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
edited Jan 3 at 10:13
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
answered Jan 3 at 10:03
drhabdrhab
98.9k544130
98.9k544130
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
add a comment |
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
$begingroup$
So if I understand correctly, because I have two distinct numbers ($ 8,9 $) I had to multiply $ {10 choose 2} $ by $ 2! $ ?
$endgroup$
– bm1125
Jan 3 at 10:15
2
2
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
$begingroup$
That is a way to put it. Personally I would rather say that there are $frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively.
$endgroup$
– drhab
Jan 3 at 10:18
add a comment |
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$begingroup$
Well, you could consider the number of integer solutions to $$ x_1 + x_2 + ldots + x_{10} = 97 $$ You can do it with the stars and bars method.
$endgroup$
– Matti P.
Jan 3 at 9:54
$begingroup$
Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation?
$endgroup$
– bm1125
Jan 3 at 9:58