Probability - The birthday problem
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You go to a party which has 1000 guests (including you).
a) What is the probability that exactly one other guest has the same birthday as you?
b) What is the probability that at least two other guests have the same birthday as you?
My proposed solutions:
a) $(frac1365)binom9991(frac364365)^998 = 0.1771$
b) Probability of zero guests or one guest NOT having the same birthday.
Zero guests: $(frac364365)^999 = 0.0645$
One guest: $0.1771$ (part a)
At least two guests: $1 - (0.1771 + 0.0645) = 0.7584$
Would this be the correct way to go about it?
probability combinations birthday
edited Nov 17 '17 at 20:45
Henry
93.5k471149
asked Sep 10 '17 at 1:30
Hello
1591217
add a comment |Â
up vote
0
down vote
favorite
You go to a party which has 1000 guests (including you).
a) What is the probability that exactly one other guest has the same birthday as you?
b) What is the probability that at least two other guests have the same birthday as you?
My proposed solutions:
a) $(frac1365)binom9991(frac364365)^998 = 0.1771$
b) Probability of zero guests or one guest NOT having the same birthday.
Zero guests: $(frac364365)^999 = 0.0645$
One guest: $0.1771$ (part a)
At least two guests: $1 - (0.1771 + 0.0645) = 0.7584$
Would this be the correct way to go about it?
probability combinations birthday
edited Nov 17 '17 at 20:45
Henry
93.5k471149
asked Sep 10 '17 at 1:30
Hello
1591217
Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
You go to a party which has 1000 guests (including you).
a) What is the probability that exactly one other guest has the same birthday as you?
b) What is the probability that at least two other guests have the same birthday as you?
My proposed solutions:
a) $(frac1365)binom9991(frac364365)^998 = 0.1771$
b) Probability of zero guests or one guest NOT having the same birthday.
Zero guests: $(frac364365)^999 = 0.0645$
One guest: $0.1771$ (part a)
At least two guests: $1 - (0.1771 + 0.0645) = 0.7584$
Would this be the correct way to go about it?
probability combinations birthday
edited Nov 17 '17 at 20:45
Henry
93.5k471149
asked Sep 10 '17 at 1:30
Hello
1591217
You go to a party which has 1000 guests (including you).
a) What is the probability that exactly one other guest has the same birthday as you?
b) What is the probability that at least two other guests have the same birthday as you?
My proposed solutions:
a) $(frac1365)binom9991(frac364365)^998 = 0.1771$
b) Probability of zero guests or one guest NOT having the same birthday.
Zero guests: $(frac364365)^999 = 0.0645$
One guest: $0.1771$ (part a)
At least two guests: $1 - (0.1771 + 0.0645) = 0.7584$
Would this be the correct way to go about it?
probability combinations birthday
edited Nov 17 '17 at 20:45
Henry
93.5k471149
asked Sep 10 '17 at 1:30
Hello
1591217
edited Nov 17 '17 at 20:45
Henry
93.5k471149
edited Nov 17 '17 at 20:45
Henry
93.5k471149
edited Nov 17 '17 at 20:45
Henry
93.5k471149
93.5k471149
asked Sep 10 '17 at 1:30
Hello
1591217
asked Sep 10 '17 at 1:30
Hello
1591217
asked Sep 10 '17 at 1:30
Hello
1591217
1591217
Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49
add a comment |Â
Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49
Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
With the standard birthday assumptions ($365$ days equally likely and each person's birthday independent of the others), your calculations look correct, as is your application of the binomial distribution
As a check in R
> dbinom(1, 999, 1/365)
[1] 0.1770821
>
> 1 - pbinom(1, 999, 1/365)
[1] 0.7583954
> sum(dbinom(2:999, 999, 1/365))
[1] 0.7583954
answered Aug 15 at 23:27
Henry
93.5k471149
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
With the standard birthday assumptions ($365$ days equally likely and each person's birthday independent of the others), your calculations look correct, as is your application of the binomial distribution
As a check in R
> dbinom(1, 999, 1/365)
[1] 0.1770821
>
> 1 - pbinom(1, 999, 1/365)
[1] 0.7583954
> sum(dbinom(2:999, 999, 1/365))
[1] 0.7583954
answered Aug 15 at 23:27
Henry
93.5k471149
add a comment |Â
up vote
1
down vote
With the standard birthday assumptions ($365$ days equally likely and each person's birthday independent of the others), your calculations look correct, as is your application of the binomial distribution
As a check in R
> dbinom(1, 999, 1/365)
[1] 0.1770821
>
> 1 - pbinom(1, 999, 1/365)
[1] 0.7583954
> sum(dbinom(2:999, 999, 1/365))
[1] 0.7583954
answered Aug 15 at 23:27
Henry
93.5k471149
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With the standard birthday assumptions ($365$ days equally likely and each person's birthday independent of the others), your calculations look correct, as is your application of the binomial distribution
As a check in R
> dbinom(1, 999, 1/365)
[1] 0.1770821
>
> 1 - pbinom(1, 999, 1/365)
[1] 0.7583954
> sum(dbinom(2:999, 999, 1/365))
[1] 0.7583954
answered Aug 15 at 23:27
Henry
93.5k471149
With the standard birthday assumptions ($365$ days equally likely and each person's birthday independent of the others), your calculations look correct, as is your application of the binomial distribution
As a check in R
> dbinom(1, 999, 1/365)
[1] 0.1770821
>
> 1 - pbinom(1, 999, 1/365)
[1] 0.7583954
> sum(dbinom(2:999, 999, 1/365))
[1] 0.7583954
answered Aug 15 at 23:27
Henry
93.5k471149
answered Aug 15 at 23:27
Henry
93.5k471149
answered Aug 15 at 23:27
Henry
93.5k471149
answered Aug 15 at 23:27
Henry
93.5k471149
93.5k471149
add a comment |Â
add a comment |Â
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Part a. looks good. Don't understand the Zero Guest formula. the probability that a given guest has a different birthday than yours is $frac 364365$ so the answer is $left( frac 364365 right)^999$. I don't understand the other factors you have.
– lulu
Sep 10 '17 at 1:33
Sorry I guess I was thinking too hard. So the zero guest would have an answer of 0.0645 for the final part b answer of 0.7584?
– Hello
Sep 10 '17 at 1:35
Should have said: your overall methodology looks fine. It's just that one calculation I have a problem with.
– lulu
Sep 10 '17 at 1:35
Yes, that's the value I get.
– lulu
Sep 10 '17 at 1:35
if the birthdays were equally distributed you could use the pigeonhole principle was my first thought.
– user451844
Sep 10 '17 at 1:49