If it is a regular dice then the probability is 3/6 that is 1/2
It refers to two event which are equally likely to occur.
The probability of you understanding this answer is slim... does that help?? That is how you use it in a sentance. ** Jeez, that's mean, and OP, check out people who have the same exact question as you on here.. "What is probability" rather than how.
Rolling the dice once will result in any one of the six numbers having the same probability of being up. The probability of getting a '5' = 1/6, the same as getting a '1.' ============================
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)!]
The probability with 30 people is 0.7063 approx.
Equiprobable, but I would stick with simplicity of communication and go with "having the same probability".
If it is a regular dice then the probability is 3/6 that is 1/2
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.
equally likely
It refers to two event which are equally likely to occur.
The probability that 2 people have the same number is 2 out of 10
Leaving aside leap years, the probability is 0.0137
The probability of you understanding this answer is slim... does that help?? That is how you use it in a sentance. ** Jeez, that's mean, and OP, check out people who have the same exact question as you on here.. "What is probability" rather than how.
0-1%
1/16?
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.