How to select distribution? — Binomial, Poisson, …
up vote
2
down vote
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How do I go about finding which distribution I need to use for my exercise?
I have the following exercise:
Compute the probability that within a group of 5 students exactly two
are born on a Sunday.
What gives me a hint on what probability distribution that is?
probability probability-theory probability-distributions
asked 10 hours ago
thebilly
284
add a comment |
up vote
2
down vote
favorite
How do I go about finding which distribution I need to use for my exercise?
I have the following exercise:
Compute the probability that within a group of 5 students exactly two
are born on a Sunday.
What gives me a hint on what probability distribution that is?
probability probability-theory probability-distributions
asked 10 hours ago
thebilly
284
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How do I go about finding which distribution I need to use for my exercise?
I have the following exercise:
Compute the probability that within a group of 5 students exactly two
are born on a Sunday.
What gives me a hint on what probability distribution that is?
probability probability-theory probability-distributions
asked 10 hours ago
thebilly
284
How do I go about finding which distribution I need to use for my exercise?
I have the following exercise:
Compute the probability that within a group of 5 students exactly two
are born on a Sunday.
What gives me a hint on what probability distribution that is?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked 10 hours ago
thebilly
284
asked 10 hours ago
thebilly
284
asked 10 hours ago
thebilly
284
asked 10 hours ago
thebilly
284
asked 10 hours ago
thebilly
284
284
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
Guide:
The total number of students is fixed. The outcome can't be exactly $6$ born on a Sunday.
Each student is either born on a Sunday or not a Sunday. We assume that they are independent.
answered 10 hours ago
Siong Thye Goh
94.5k1462114
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
add a comment |
up vote
1
down vote
The Guide from the previous answer is helpful. It should help you pinpoint the exact distribution to choose.
Sometimes the problem might not be easily convertible to a distribution. In those cases you can always go back to thinking in terms of basic probabilities.
How many total ways can the student birthday be arranged? Number of days to the power of number of students = $7^5$
How many ways can we arrange the students to satisfy the condition? Choose 2 students from the 5. They are born on a Sunday. The rest isn't. This leads to ${5 choose 2}$ * $1^2$ * $6^3$
If you divide the two you will get ${5 choose 2} * {frac 1 7}^2 * {frac 6 7}^3$
Notice that it's the same result if you were to model it using binomial distribution with p = 1/7, n = 5. So at the end of the day, if you feel stuck you can always roll back to basics.
Hope it helps
answered 10 hours ago
Ofya
815
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Guide:
The total number of students is fixed. The outcome can't be exactly $6$ born on a Sunday.
Each student is either born on a Sunday or not a Sunday. We assume that they are independent.
answered 10 hours ago
Siong Thye Goh
94.5k1462114
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
add a comment |
up vote
5
down vote
Guide:
The total number of students is fixed. The outcome can't be exactly $6$ born on a Sunday.
Each student is either born on a Sunday or not a Sunday. We assume that they are independent.
answered 10 hours ago
Siong Thye Goh
94.5k1462114
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Guide:
The total number of students is fixed. The outcome can't be exactly $6$ born on a Sunday.
Each student is either born on a Sunday or not a Sunday. We assume that they are independent.
answered 10 hours ago
Siong Thye Goh
94.5k1462114
Guide:
The total number of students is fixed. The outcome can't be exactly $6$ born on a Sunday.
Each student is either born on a Sunday or not a Sunday. We assume that they are independent.
answered 10 hours ago
Siong Thye Goh
94.5k1462114
answered 10 hours ago
Siong Thye Goh
94.5k1462114
answered 10 hours ago
Siong Thye Goh
94.5k1462114
answered 10 hours ago
Siong Thye Goh
94.5k1462114
94.5k1462114
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
add a comment |
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
A somewhat notable detail is that, in the real world, the probability of a given person having been born on a Sunday is not $frac17$. The exact probability varies depending on where the person lives and how old they are, but in many places it is nowadays closer to $frac1{10}$. The main cause this is effect is apparently the fact that induced and caesarean births tend to be mostly scheduled for weekdays, when more hospital staff is available.
– Ilmari Karonen
10 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
7 hours ago
add a comment |
up vote
1
down vote
The Guide from the previous answer is helpful. It should help you pinpoint the exact distribution to choose.
Sometimes the problem might not be easily convertible to a distribution. In those cases you can always go back to thinking in terms of basic probabilities.
How many total ways can the student birthday be arranged? Number of days to the power of number of students = $7^5$
How many ways can we arrange the students to satisfy the condition? Choose 2 students from the 5. They are born on a Sunday. The rest isn't. This leads to ${5 choose 2}$ * $1^2$ * $6^3$
If you divide the two you will get ${5 choose 2} * {frac 1 7}^2 * {frac 6 7}^3$
Notice that it's the same result if you were to model it using binomial distribution with p = 1/7, n = 5. So at the end of the day, if you feel stuck you can always roll back to basics.
Hope it helps
answered 10 hours ago
Ofya
815
add a comment |
up vote
1
down vote
The Guide from the previous answer is helpful. It should help you pinpoint the exact distribution to choose.
Sometimes the problem might not be easily convertible to a distribution. In those cases you can always go back to thinking in terms of basic probabilities.
How many total ways can the student birthday be arranged? Number of days to the power of number of students = $7^5$
How many ways can we arrange the students to satisfy the condition? Choose 2 students from the 5. They are born on a Sunday. The rest isn't. This leads to ${5 choose 2}$ * $1^2$ * $6^3$
If you divide the two you will get ${5 choose 2} * {frac 1 7}^2 * {frac 6 7}^3$
Notice that it's the same result if you were to model it using binomial distribution with p = 1/7, n = 5. So at the end of the day, if you feel stuck you can always roll back to basics.
Hope it helps
answered 10 hours ago
Ofya
815
add a comment |
up vote
1
down vote
up vote
1
down vote
The Guide from the previous answer is helpful. It should help you pinpoint the exact distribution to choose.
Sometimes the problem might not be easily convertible to a distribution. In those cases you can always go back to thinking in terms of basic probabilities.
How many total ways can the student birthday be arranged? Number of days to the power of number of students = $7^5$
How many ways can we arrange the students to satisfy the condition? Choose 2 students from the 5. They are born on a Sunday. The rest isn't. This leads to ${5 choose 2}$ * $1^2$ * $6^3$
If you divide the two you will get ${5 choose 2} * {frac 1 7}^2 * {frac 6 7}^3$
Notice that it's the same result if you were to model it using binomial distribution with p = 1/7, n = 5. So at the end of the day, if you feel stuck you can always roll back to basics.
Hope it helps
answered 10 hours ago
Ofya
815
The Guide from the previous answer is helpful. It should help you pinpoint the exact distribution to choose.
Sometimes the problem might not be easily convertible to a distribution. In those cases you can always go back to thinking in terms of basic probabilities.
How many total ways can the student birthday be arranged? Number of days to the power of number of students = $7^5$
How many ways can we arrange the students to satisfy the condition? Choose 2 students from the 5. They are born on a Sunday. The rest isn't. This leads to ${5 choose 2}$ * $1^2$ * $6^3$
If you divide the two you will get ${5 choose 2} * {frac 1 7}^2 * {frac 6 7}^3$
Notice that it's the same result if you were to model it using binomial distribution with p = 1/7, n = 5. So at the end of the day, if you feel stuck you can always roll back to basics.
Hope it helps
answered 10 hours ago
Ofya
815
answered 10 hours ago
Ofya
815
answered 10 hours ago
Ofya
815
answered 10 hours ago
Ofya
815
815
add a comment |
add a comment |
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