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If 15 strangers are all in a room what is the probability of them all having the same birthday?
Wiki User
∙ 14y ago
To determine the probability of 15 random people all having the same birthday, consider each person one at a time. (This is for the non leap-year case.)
The probability of any person having any birthday is 365 in 365, or 1.
The probability of any other person having that same birthday is 1 in 365, or 0.00274.
The probability, then, of 15 random people having the same birthday is the product of these probabilities, or 0.0027414 times 1, or 1.34x10-36.
Note: This answer assumes also that the distribution of birthdays for a large group of people in uniformly random over the 365 days of the year. That is probably not actually true. There are several non-random points of conception, some of which are spring, Valentine's day, and Christmas, depending of culture and religion. That makes the point of birth, nine months later, also be non-uniform, so that can skew the results.
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∙ 14y ago
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What is the smallest number of people in a room where the probability of two of them having the same birthday is at least 50 percent?
23. The probability that at least two people in a room share a birthday can be expressed more simply, mathematically, as 1 minus the probability that nobody in the room shares a birthday.Imagine a fairly simple example of a room with only three people. The probability that any two share a birthday is :1 - [ 365/365 x 364/365 x 363/365]i.e. 1-P(none of them share a birthday)=1 - [ (365x364x363) / 3653 ]=0.8%Similarly,P(any two share a birthday in a room of 4 people)= 1 - [ 365x364x363x362 / 3654 ] = 1.6%If you keep following that logic eventually you getP(any two share a birthday in a room of 23 people)=1 - [(365x364x...x344x343) / 36523 ] = 51%
What is the probability that in a room of 8 people 2 have the same birthday?
P = 0.072314699... ≈ 7.23%SOLUTIONLet's identify the 8 people by integers 1, 2,..., 8, and their birthday by x1, x2,..., x8.To include the possibility of xi = Feb. 29 (of a leap year), we will consider 4(365) +1 = 1451 days, so the probability of a person having Feb. 29 as his birthday isP(xi = Feb. 29) = 1/1461. The probability of a person having his birthday on anygiven day except Feb. 29 is P(xi ≠ Feb. 29) = 4/1461.Let's first calculate the probability of person 1 and 2 having the same birthday, x1 = x2, and persons 3, 4,..., 8, all have different birthdays x3,...,x8.Now we go.We take person 1 and have two possibilities:A).- x1 = Feb. 29 B).- x1 ≠ Feb. 29P(x1 = Feb. 29) = 1/1461; P(x ≠ Feb. 29) = 1 - 1/1461We will construct two branches. One for each possibility.Branch A: given x1 = Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x1) = 1 - 1/1461. The probability of person 3 not having the same birthday is P(x3 ≠ x1 = x2) = 1 - 1/1461. The probability of person 4 not having either two birthdays is P(x4 ≠ x1, x4 ≠ x3) = 1 - 5/1461. The probability of person 5 not sharing x1 nor x3 nor x4 is P(x5 ≠ x4, x5 ≠ x3, x5 ≠ x2) = 1 - 9/1461. And so a series of terms develops:P(xi ≠ ... ) = 1 - (4i - 3)/1461 from i = 1, to i = n-2where n is the number of people in the room (In this particular question for 8people, n = 8).The branch we have constructed for possibility A is the product of the individualconditional probabilities;A: (1/1461)2 π1n-2[1-(4i-3)/1461]where π1n-2 is the product operator "pi" of "i" terms from i =1, to i = n-2.Branch B: Given x1 ≠Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x2) = 4/1461. The probability of person 3 not having the same birthdayis P(x3 ≠ x1) = 1 - 4/1461. The probability of person 4 not sharing birthdays with 2 and 3 is P(x4 ≠ x3, x4 ≠ x2) = 1 - 8/1451. The probability of person 5 notsharing x2 nor x3 nor x4 is P(x5 ≠ ...) = 1 - 12/1461. And so a series of termsdevelops:P(xi ≠ ... ) = 1- (4i -4)/1461 from i = 2, to i = n-1where n is the number of people in the room (In this particular question for 8 people, n = 8).Branch for possibility B: (1-1/1461)(4/1461) π2n-1[1-(4i-4)/1461]The sum of these two branch expressions give the probability of persons 1 and 2sharing a birthday in a group of n persons:P(x1=x2)=(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]Now, the probability of any two persons, and only those two persons, to share a birthday in a group of n persons, is the previous probability, times the number of different combinations of two persons, (i,j), in a group of n persons, nC2(where, nC2 = n!/[2!(n-2)!] )The final expression is:--------------------------------------------------------------------------------------------P = nC2{(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]}--------------------------------------------------------------------------------------------For the case of n = 8, the result is, P = 0.072314699... ≈ 7.23%
What is the largest possible number of people in a room if no two people have a birthday in the same month?
12 because there are 12 months and no one in the room has a birthday in the same month
If there are 10 people in a room what is the probability that any 2 people have the same birthday?
11.7%. (But when there are 23 people, the odds are 50.7%.) In order for 2 people to have the same birthday, then there can only be 9 different dates out of 365 (actually 37 out of 1461). This is roughly equivalent to 2.53% of possible dates. However, the chance of having matching birthdays to any other birthday is actually much larger, because EACH individual has up to 9 chances out of 365 to match another. The probability is calculated using the product of the series p(n) = 1-(n-1/365), which compares the likelihood of all possible pairs of identical birthdays. The chance that no two share a birthday (including the leap day) can be calculated as the product of the terms n = 1 to 10 of [(1461-(4n+1)/1461]. This should also equate to a likelihood of 88% because the Leap Day is only 1 chance in 1461. (see related link)
On average how many people share the same birthday?
On average, in the entire world about twenty eight thousand people will have the same birthday as one another. In a room full of people, there should be two people who have the same birthday.
What is the smallest number of people in a room where the probability of two of them having the same birthday is at least 50 percent?
23. The probability that at least two people in a room share a birthday can be expressed more simply, mathematically, as 1 minus the probability that nobody in the room shares a birthday.Imagine a fairly simple example of a room with only three people. The probability that any two share a birthday is :1 - [ 365/365 x 364/365 x 363/365]i.e. 1-P(none of them share a birthday)=1 - [ (365x364x363) / 3653 ]=0.8%Similarly,P(any two share a birthday in a room of 4 people)= 1 - [ 365x364x363x362 / 3654 ] = 1.6%If you keep following that logic eventually you getP(any two share a birthday in a room of 23 people)=1 - [(365x364x...x344x343) / 36523 ] = 51%
What is the probability of one person in a random group of fifty people having a birthday today?
About a 12.8 percent chance. The math is actually very simple: q(n)= 1- (364/365)n Where "n" is the number of people present. It is worth noting that in a room of 50 people, there is a 97% chance that two of them share the same birthday.
Virtual party room code on Build-a-Bearville?
You can only get a virtual party room by having a birthday party at your local Build-A-Bear Workshop store.
What is the formula for calculating the birthday paradox?
In a room of just 23 people there's a 50-50 chance of two people having the same birthday. In a room of 75 there's a 99.9% chance of two people matching birth dates.
What is the probability that two people selected from 30 have the same birthday?
To select the first birthday, the probability is 1/30. Having gotten that, the conditional probability that the next birthday would be the same is (1/30)x(1/29) and that is 1/870----------------------------------------------------------------------------------------------I believe the question has to be rephrased to "What is the probability that two peoplein a group of 30 people share the same birthday?". Because in the way the questionis actually stated, "the probability that two persons selected randomly from a group of 30 have the same birthday", the event that "those two people would share theirbirthday" is independent of the size of the population they were selected from.In the case the actual question be "What is the probability that two people in a groupof 30 share the same birthday?, is given by the following expression that neglectsFebruary 29 (of the leap), but gives very good approximation to the expression thatconsiders February 29 and is a simpler one. [It has to be mentioned that the analysisleading to this expression considers birthdays a "random variable" where chances fora persons birthday are the same for any day of the year]:P(2 share bd out of n) = nC2 (1/365) Π1n-1[1-(i-1)/365]for n = 30, P(2 share bd out of 30) = 30C2 (1/365) Π1 29 [1-(i-1)/365] = 435∙(1/365)∙[1-1/365]∙[1-2/365]∙[1-3/365]∙ ∙∙∙ ∙[1-28/365] = 0.380215577... ≈ 0.380 ≈ 38.0%For the construction of the expression to calculate the probability of any two people sharing a birthday in a group of n people considering Feb 29 of the leap year see thequestion "What is the probability that in a room of 8 people 2 have the same birthday?"
Does great wolf lodge give out birthday presents if your having a birthday party there?
The great wolf lodge has birthday packages, you can find them when you book a room, but other than that no. And you have to pay for the packages. So no the Great Wolf Lodge does not give out free birthday party gifts.
Will carbon monoxide catch fire in a heated utility room?
The probability is not significant.
When Bella is having her baby are her parents in the room?
Bella's parents are not in the room when she is having her baby.
If a client has oxygen what is more dangerous smoking in the room or having oils in the room?
Having oils in the room is more dangerous.
What is the probability that in a room of 8 people 2 have the same birthday?
P = 0.072314699... ≈ 7.23%SOLUTIONLet's identify the 8 people by integers 1, 2,..., 8, and their birthday by x1, x2,..., x8.To include the possibility of xi = Feb. 29 (of a leap year), we will consider 4(365) +1 = 1451 days, so the probability of a person having Feb. 29 as his birthday isP(xi = Feb. 29) = 1/1461. The probability of a person having his birthday on anygiven day except Feb. 29 is P(xi ≠ Feb. 29) = 4/1461.Let's first calculate the probability of person 1 and 2 having the same birthday, x1 = x2, and persons 3, 4,..., 8, all have different birthdays x3,...,x8.Now we go.We take person 1 and have two possibilities:A).- x1 = Feb. 29 B).- x1 ≠ Feb. 29P(x1 = Feb. 29) = 1/1461; P(x ≠ Feb. 29) = 1 - 1/1461We will construct two branches. One for each possibility.Branch A: given x1 = Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x1) = 1 - 1/1461. The probability of person 3 not having the same birthday is P(x3 ≠ x1 = x2) = 1 - 1/1461. The probability of person 4 not having either two birthdays is P(x4 ≠ x1, x4 ≠ x3) = 1 - 5/1461. The probability of person 5 not sharing x1 nor x3 nor x4 is P(x5 ≠ x4, x5 ≠ x3, x5 ≠ x2) = 1 - 9/1461. And so a series of terms develops:P(xi ≠ ... ) = 1 - (4i - 3)/1461 from i = 1, to i = n-2where n is the number of people in the room (In this particular question for 8people, n = 8).The branch we have constructed for possibility A is the product of the individualconditional probabilities;A: (1/1461)2 π1n-2[1-(4i-3)/1461]where π1n-2 is the product operator "pi" of "i" terms from i =1, to i = n-2.Branch B: Given x1 ≠Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x2) = 4/1461. The probability of person 3 not having the same birthdayis P(x3 ≠ x1) = 1 - 4/1461. The probability of person 4 not sharing birthdays with 2 and 3 is P(x4 ≠ x3, x4 ≠ x2) = 1 - 8/1451. The probability of person 5 notsharing x2 nor x3 nor x4 is P(x5 ≠ ...) = 1 - 12/1461. And so a series of termsdevelops:P(xi ≠ ... ) = 1- (4i -4)/1461 from i = 2, to i = n-1where n is the number of people in the room (In this particular question for 8 people, n = 8).Branch for possibility B: (1-1/1461)(4/1461) π2n-1[1-(4i-4)/1461]The sum of these two branch expressions give the probability of persons 1 and 2sharing a birthday in a group of n persons:P(x1=x2)=(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]Now, the probability of any two persons, and only those two persons, to share a birthday in a group of n persons, is the previous probability, times the number of different combinations of two persons, (i,j), in a group of n persons, nC2(where, nC2 = n!/[2!(n-2)!] )The final expression is:--------------------------------------------------------------------------------------------P = nC2{(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]}--------------------------------------------------------------------------------------------For the case of n = 8, the result is, P = 0.072314699... ≈ 7.23%
What can you do for an 11th birthday party in a hotel?
Is there a swimming pool or work out area? Perhaps the celebrant is interested in having a some friends meet there. Specialty restaurant, game room? Hire a function room at the hotel of an appropriate size. Music and party food.
What is the largest possible number of people in a room if no two people have a birthday in the same month?
12 because there are 12 months and no one in the room has a birthday in the same month
