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What is probability that 2 persons sharing same birth date?
The probability that two persons share the same birth date can be calculated using the concept of the birthday paradox. In a group of 23 people, there is a probability of approximately 50% that two individuals share the same birth date. This probability increases as the number of people in the group increases due to the increasing number of possible pairs to compare. The calculation involves considering the complementary probability of no one sharing a birthday and subtracting it from 1 to find the probability of at least one shared birthday.
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
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.
A nursing class has 13 women and 18 men if a student chosen randomly to be the team leader what is the probability the student is a woman?
least likely
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.
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 probability that 2 persons sharing same birth date?
The probability that two persons share the same birth date can be calculated using the concept of the birthday paradox. In a group of 23 people, there is a probability of approximately 50% that two individuals share the same birth date. This probability increases as the number of people in the group increases due to the increasing number of possible pairs to compare. The calculation involves considering the complementary probability of no one sharing a birthday and subtracting it from 1 to find the probability of at least one shared 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 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 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 is the probability that at least 2 students in a class of 36 have the same birthday?
About 83.2%The probability that non of the 36 students have the same birthday (not consideringFebruary 28 of the leap year) is given by the following relation:P(non out of n have same bd) = Π1n-1 [(365-i)/365]P(non out of 36 have same bd) = (364/365)(363/365)(362/365) ... (331/365)(330/365) == 0.167817892.. ≈ 16.8%So the probability of at least 2 having the same birthday is about 1 - .168 = 0.832 =83.2%
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
If 20 percent of people who enter store buy something and 3 people enter what is the probability of at least one sale?
The probability of at least one event occurring out of several events is equal to one minus the probability of none of the events occurring. This is a binomial probability problem. Go to any binomial probability table with p=0.2, n=3 and the probability of 0 is 0.512. Therefore, 1-0.512 is 0.488 which is the probability of at least 1 sale.
Each of 3 people chose a number from 1-10 what is the probability that at least 2 people have the same number?
The probability that 2 people have the same number is 2 out of 10
The probability that an individual is left handed is 0.1 In a class of 30 students what is the probability of finding at least 5 left hander's?
This is a binomial probability distribution. The number of trials, n, equals 30; and the probability of success is p, which is 0.1. In this problem, you want the probability of at least 5, which is the complement of at most 4. We use the complement because we can subtract from 1 that probability and we will have the solution. The related link has the binomial probability distribution table which is cumulative. Per the table, at n=30, p=0.1 and x = 4; the probability is 0.825. Therefore the probability of at least 5 is 1 - 0.825 or 0.175.
What is a consumer class?
A class of People who earn at least 7,000 a year.
