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 Nov 28 at 16:05
thebilly
565
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 Nov 28 at 16:05
thebilly
565
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 Nov 28 at 16:05
thebilly
565
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 Nov 28 at 16:05
thebilly
565
asked Nov 28 at 16:05
thebilly
565
asked Nov 28 at 16:05
thebilly
565
asked Nov 28 at 16:05
thebilly
565
asked Nov 28 at 16:05
thebilly
565
565
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 Nov 28 at 16:12
Siong Thye Goh
98k1463116
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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 Nov 28 at 16:26
Ofya
4948
add a comment |
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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 Nov 28 at 16:12
Siong Thye Goh
98k1463116
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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 Nov 28 at 16:12
Siong Thye Goh
98k1463116
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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 Nov 28 at 16:12
Siong Thye Goh
98k1463116
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 Nov 28 at 16:12
Siong Thye Goh
98k1463116
answered Nov 28 at 16:12
Siong Thye Goh
98k1463116
answered Nov 28 at 16:12
Siong Thye Goh
98k1463116
answered Nov 28 at 16:12
Siong Thye Goh
98k1463116
98k1463116
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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
Nov 28 at 16:51
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
Nov 28 at 16:51
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
Wow. ok. now I feel stupid. seems to be a binomial.. either it is a success (born on Sunday) or not... Thank you.
– thebilly
Nov 28 at 19:18
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 Nov 28 at 16:26
Ofya
4948
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 Nov 28 at 16:26
Ofya
4948
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 Nov 28 at 16:26
Ofya
4948
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 Nov 28 at 16:26
Ofya
4948
answered Nov 28 at 16:26
Ofya
4948
answered Nov 28 at 16:26
Ofya
4948
answered Nov 28 at 16:26
Ofya
4948
4948
add a comment |
add a comment |
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