Compute $ P{X_1 = 3 | X = 10 } $, $ X sim Pois(10) $
$begingroup$
Suppose $ X$ is the number of people entering a store in an hour and $ X sim Pois(10) $. Compute the probability that at most 3 men enter the store if it is known that 10 women already entered. What assumptions do you make?
I guess the assumption need to make is that the chance of a female male entering the store is the same $ (0.5) $ and therefore the number of men entering the store is also poisson distribution with $ lambda = 5 $ ?
Not sure how to proceed from here (if it is true)? Lets say $ X_1 $ is number of men entering the store (with the values $ 0,1,2,3 $ ) and $ X_2 $ number of women.
$ P{X_1 = 3 | X_2 = 10 } = frac{P{X_1 = 3, X_2 = 10 }}{P{X_2 = 10 }} $
probability poisson-distribution
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
$endgroup$
add a comment |
$begingroup$
Suppose $ X$ is the number of people entering a store in an hour and $ X sim Pois(10) $. Compute the probability that at most 3 men enter the store if it is known that 10 women already entered. What assumptions do you make?
I guess the assumption need to make is that the chance of a female male entering the store is the same $ (0.5) $ and therefore the number of men entering the store is also poisson distribution with $ lambda = 5 $ ?
Not sure how to proceed from here (if it is true)? Lets say $ X_1 $ is number of men entering the store (with the values $ 0,1,2,3 $ ) and $ X_2 $ number of women.
$ P{X_1 = 3 | X_2 = 10 } = frac{P{X_1 = 3, X_2 = 10 }}{P{X_2 = 10 }} $
probability poisson-distribution
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
$endgroup$
add a comment |
$begingroup$
Suppose $ X$ is the number of people entering a store in an hour and $ X sim Pois(10) $. Compute the probability that at most 3 men enter the store if it is known that 10 women already entered. What assumptions do you make?
I guess the assumption need to make is that the chance of a female male entering the store is the same $ (0.5) $ and therefore the number of men entering the store is also poisson distribution with $ lambda = 5 $ ?
Not sure how to proceed from here (if it is true)? Lets say $ X_1 $ is number of men entering the store (with the values $ 0,1,2,3 $ ) and $ X_2 $ number of women.
$ P{X_1 = 3 | X_2 = 10 } = frac{P{X_1 = 3, X_2 = 10 }}{P{X_2 = 10 }} $
probability poisson-distribution
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
$endgroup$
Suppose $ X$ is the number of people entering a store in an hour and $ X sim Pois(10) $. Compute the probability that at most 3 men enter the store if it is known that 10 women already entered. What assumptions do you make?
I guess the assumption need to make is that the chance of a female male entering the store is the same $ (0.5) $ and therefore the number of men entering the store is also poisson distribution with $ lambda = 5 $ ?
Not sure how to proceed from here (if it is true)? Lets say $ X_1 $ is number of men entering the store (with the values $ 0,1,2,3 $ ) and $ X_2 $ number of women.
$ P{X_1 = 3 | X_2 = 10 } = frac{P{X_1 = 3, X_2 = 10 }}{P{X_2 = 10 }} $
probability poisson-distribution
probability poisson-distribution
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
asked Dec 31 '18 at 13:15
bm1125bm1125
67816
67816
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $lambda=5$.
The independence tells us that: $$P(X_1leq3mid X_2=10)=P(X_1leq3)$$
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
$endgroup$
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $lambda=5$.
The independence tells us that: $$P(X_1leq3mid X_2=10)=P(X_1leq3)$$
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
$endgroup$
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
add a comment |
$begingroup$
Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $lambda=5$.
The independence tells us that: $$P(X_1leq3mid X_2=10)=P(X_1leq3)$$
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
$endgroup$
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
add a comment |
$begingroup$
Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $lambda=5$.
The independence tells us that: $$P(X_1leq3mid X_2=10)=P(X_1leq3)$$
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
$endgroup$
Since further info lacks you are entitled to go for the following:
$X=X_1+X_2$ where $X_1,X_2$ are iid and have Poisson distribution with rate $lambda=5$.
The independence tells us that: $$P(X_1leq3mid X_2=10)=P(X_1leq3)$$
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
answered Dec 31 '18 at 13:43
drhabdrhab
103k545136
103k545136
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
add a comment |
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
You mean $ X_1 , X_2 $ are the number of the men and women entering the store with $ lambda = 5 $ each??
$endgroup$
– bm1125
Dec 31 '18 at 14:08
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
$begingroup$
Yes, that is what I mean. Then $X=X_1+X_2$ has Poisson distribution with rate $10$. The number of men entering the shop will not depend on the number of women (unless they come by e.g. pairs, but no info is given about that).
$endgroup$
– drhab
Dec 31 '18 at 15:02
add a comment |
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