Poisson Process conditional probability
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0
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I have the problem
"Suppose that the number of calls arriving to a telesales call centre follows
a Poisson process with intensity λ = 6 calls per hour. Each caller makes a
purchase with probability 0.4.
Suppose now that 5 calls arrived in the first hour. What is the probability
that 3 of them made a purchase?"
I have set $X_t =$ number of calls by time $ t $ and $Y_t =$ number of purchases made by time $ t $
So I have $Y_t sim Po(0.4(6)t)$ and I found that arrival time is exponentially distributed and from that i have the expected arrival $mu = frac{1}{2.4}=0.4167$
I then found $$P(Y_1=3)= 0.20914 $$
Im not really sure how to use all this information to get the answer I want, I assume the answer is $$P(Y_1=3|X_1=5)$$ but not sure what to do
probability stochastic-processes poisson-process
asked Nov 22 at 11:35
Rito Lowe
465
add a comment |
up vote
0
down vote
favorite
I have the problem
"Suppose that the number of calls arriving to a telesales call centre follows
a Poisson process with intensity λ = 6 calls per hour. Each caller makes a
purchase with probability 0.4.
Suppose now that 5 calls arrived in the first hour. What is the probability
that 3 of them made a purchase?"
I have set $X_t =$ number of calls by time $ t $ and $Y_t =$ number of purchases made by time $ t $
So I have $Y_t sim Po(0.4(6)t)$ and I found that arrival time is exponentially distributed and from that i have the expected arrival $mu = frac{1}{2.4}=0.4167$
I then found $$P(Y_1=3)= 0.20914 $$
Im not really sure how to use all this information to get the answer I want, I assume the answer is $$P(Y_1=3|X_1=5)$$ but not sure what to do
probability stochastic-processes poisson-process
asked Nov 22 at 11:35
Rito Lowe
465
In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the problem
"Suppose that the number of calls arriving to a telesales call centre follows
a Poisson process with intensity λ = 6 calls per hour. Each caller makes a
purchase with probability 0.4.
Suppose now that 5 calls arrived in the first hour. What is the probability
that 3 of them made a purchase?"
I have set $X_t =$ number of calls by time $ t $ and $Y_t =$ number of purchases made by time $ t $
So I have $Y_t sim Po(0.4(6)t)$ and I found that arrival time is exponentially distributed and from that i have the expected arrival $mu = frac{1}{2.4}=0.4167$
I then found $$P(Y_1=3)= 0.20914 $$
Im not really sure how to use all this information to get the answer I want, I assume the answer is $$P(Y_1=3|X_1=5)$$ but not sure what to do
probability stochastic-processes poisson-process
asked Nov 22 at 11:35
Rito Lowe
465
I have the problem
"Suppose that the number of calls arriving to a telesales call centre follows
a Poisson process with intensity λ = 6 calls per hour. Each caller makes a
purchase with probability 0.4.
Suppose now that 5 calls arrived in the first hour. What is the probability
that 3 of them made a purchase?"
I have set $X_t =$ number of calls by time $ t $ and $Y_t =$ number of purchases made by time $ t $
So I have $Y_t sim Po(0.4(6)t)$ and I found that arrival time is exponentially distributed and from that i have the expected arrival $mu = frac{1}{2.4}=0.4167$
I then found $$P(Y_1=3)= 0.20914 $$
Im not really sure how to use all this information to get the answer I want, I assume the answer is $$P(Y_1=3|X_1=5)$$ but not sure what to do
probability stochastic-processes poisson-process
probability stochastic-processes poisson-process
asked Nov 22 at 11:35
Rito Lowe
465
asked Nov 22 at 11:35
Rito Lowe
465
edited Nov 23 at 12:37
asked Nov 22 at 11:35
Rito Lowe
465
asked Nov 22 at 11:35
Rito Lowe
465
asked Nov 22 at 11:35
Rito Lowe
465
465
In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01
add a comment |
In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01
In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01
In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01
add a comment |
1 Answer
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oldest
votes
up vote
2
down vote
accepted
Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.
answered Nov 22 at 11:53
3684
1277
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.
answered Nov 22 at 11:53
3684
1277
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
add a comment |
up vote
2
down vote
accepted
Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.
answered Nov 22 at 11:53
3684
1277
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.
answered Nov 22 at 11:53
3684
1277
Since you already know that $5$ people called you no longer need to use the Poisson distribution. Try answer this question: "$5$ calls arrived in the first hour. Giver that each caller has a $0.4$ probability to make a purchase, what is the probability that $3$ of them made a purchase." If you are still unable to do it I am happy to provide a full solution.
answered Nov 22 at 11:53
3684
1277
answered Nov 22 at 11:53
3684
1277
answered Nov 22 at 11:53
3684
1277
answered Nov 22 at 11:53
3684
1277
1277
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
add a comment |
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
ok so I would use cumulative binomial probabilities? $ P(Y=3) = ({5choose3})(0.4)^{3}(0.6)^{2}$ is this correct
– Rito Lowe
Nov 26 at 18:42
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
Yes. Do you understand why it is ${5choose3}(0.4)^3(0.6)^2$?
– 3684
Nov 27 at 12:19
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
yes, because ${5choose3}$ is the number of different ways in which 3 purchases can occur within the 5 calls
– Rito Lowe
Nov 28 at 16:07
That's correct.
– 3684
Nov 28 at 23:17
That's correct.
– 3684
Nov 28 at 23:17
add a comment |
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In your last expression $t$ should be replaced by $1$.
– drhab
Nov 22 at 12:01