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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 get
P(any two share a birthday in a room of 23 people)
=1 - [(365x364x...x344x343) / 36523 ] = 51%
What else can I help you with?
What is the probability of 2 or more people in a a group of about 30 having the same birthday?
The probability with 30 people is 0.7063 approx.
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.
If the probability of a defect is 0.015 what is the probability of not having a defect?
1-.015 = .985
What is the probability of having 3 baby girls in a row?
Assuming that the probability of having a baby girl is 1/2 and that of having a baby boy is 1/2, the probability of having 3 baby girls in a row is (1/2)(1/2)(1/2)=1/8.
What is the Probability of having a female offspring?
50%
What is the probability of 2 or more people in a a group of about 30 having the same birthday?
The probability with 30 people is 0.7063 approx.
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.
Probability of getting married and having children?
A man has an 81 percent chance to get married if they live the United States before the age of 40. A woman has an 86 percent chance of getting married.
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 two black guinea pigs having black offspring?
depends on the two guinea pigs genotypes. could be anywhere from 75 to 100 percent.
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.
If the probability of a defect is 0.015 what is the probability of not having a defect?
1-.015 = .985
What is the probability of having a son?
50%
What mathematical term is used in having the same probability?
Equiprobable, but I would stick with simplicity of communication and go with "having the same probability".
What is the probability that out of 40 students none of them will have a birthday in March?
3 % No, not correct. ------------------------------------------------------------------------------------------------ The probability that a single person would have a birthday in March is 1 out of 12 (because there are 12 months in the year). Hence the probability that one of 40 students would have a birthday in March is 40 x 1/12 = 10/3 = 33.33%; More accurately March has 31 days out of 365 days min the year so the probability of one person having a birthday in March is 31/365, and for 40 students it would be 4 x 31/365 = 124/365 = 0.3397(to 4 decimal places) = 34% to nearest 1%
Why can't there be probability density equal to 100 percent in an orbital?
The probability density in an orbital cannot be equal to 100 percent because an electron exhibits wave-like behavior in quantum mechanics, meaning it does not have a definite position. Instead, the probability density provides the likelihood of finding an electron in a particular region of space within the orbital. Having a probability density of 100 percent would imply that the electron's position is known precisely, which contradicts the principles of quantum mechanics.
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)!]
