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How likely is it that in a group of 15 people who don't know each other at least 2 share a birthday?
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∙ 9y ago
10^%
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How many people must be in the room in order to guaratnee that at least two people share a birthday?
367
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 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)!]
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
How many people are needed in a room before two of them share the same birthday?
For the chance to be at least 50% that two people share the same birthday, there needs to be 22 people. For the chance to be exactly 100% that two people share the same birthday, there needs to be 366 people. If there was 365 people, there would be a very small chance that each person in the room would have different birthdays. With 366 people, there are not enough individual days for every person to have a different birthday, so there has to be at least one pair.
What is the probabililty of at least 2 people same birthday from a group of 12 people?
The probability of at least 2 people sharing a birthday in a group of 12 is approximately 0.891. This is calculated using the complement rule, finding the probability that no one shares a birthday and subtracting it from 1. The result indicates that it is highly likely for at least 2 people to share a birthday in a group of 12.
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The group least likely to be in combination with the White group was the Black group. Historically, there have been significant social, economic, and political barriers that have separated the White and Black communities in various contexts.
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1-365/[(365-6)*365^6] = 1 Is this O.K ?
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