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What is the probability of 2 or more people in a a group of about 30 having the same birthday?
Wiki User
∙ 9y ago
The probability with 30 people is 0.7063 approx.
Wiki User
∙ 9y ago
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If 15 strangers are all in a room what is the probability of them all having the same birthday?
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.
What is the probability of 26 people in a group having the same birthday?
Birthdays are not uniformly distributed over the year. But, if you do assume that they are, then: ignoring leap years, it is approx 0.5982. If you include leap years it is 0.5687.
What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
What is the probability that at least 2 people in a random group of 6 people have a birthday on the same day of the week?
1-365/[(365-6)*365^6] = 1 Is this O.K ?
What is the probability that at least 2 people have the same birth month in a group of 8 people?
Birth months are not uniformly distributed across the year. However, if yo assume that they are, the probability is 0.9536 (approx).
What is the probabililty of at least 2 people same birthday from a group of 12 people?
The probability of at least 2 people in a group of npeople sharing a common birthday can be expressed more easily (mathematically) as 1 minus the probability that nobody in the group shares a birthday. Consider two people. The probability that they don't have a common birthday is 365/365 x 364/365. So the probability that they do share a birthday is 1-(365/365 x 364/365) = 1-365x364/3652 Now consider 3 people. The probability that at least 2 share a common birthday is 1-365x364x363/3653 And so on so that the probability that at least 2 people in a group of n people having the same birthday = 1-(365x363x363x...x365-n+1)/365n = 1-365!/[ (365-n)! x 365n ]In the case of 12 people this equates to 0.16702 (or 16.7%).
What is the probabililty of at least 2 people same birthday from a group of 13 people?
19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]
If 15 strangers are all in a room what is the probability of them all having the same birthday?
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.
What are the odds of having kids with the same birthday but a different year?
In probability theory, the birthday problem, or birthday paradox[1] pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of 10 randomly chosen people, there is an 11.7% chance. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 367 (there are a maximum of 366 possible birthdays). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack. See Wikipedia for more: http://en.wikipedia.org/wiki/Birthday_paradox
What is the probability of 26 people in a group having the same birthday?
Birthdays are not uniformly distributed over the year. But, if you do assume that they are, then: ignoring leap years, it is approx 0.5982. If you include leap years it is 0.5687.
What is the probability of you and 2 complete random people sharing a birthday?
Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082
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.
What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
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?"
What is the probability that at least 2 people in a random group of 6 people have a birthday on the same day of the week?
1-365/[(365-6)*365^6] = 1 Is this O.K ?
What is the probabililty of at least 2 people same birthday from a group of n people then how large ned n for the probability to be greater than 0.5?
It is 1 - 365Cn/365n. This is greater than 0.5 for n greater than or equal to 23.
What is a sex group?
A group of people having sex
