Let x be the number of people:
P(among x, at least two people have same birthday) = 2/x * 1/(x-1) >0.6
2/[(x)x-1] = 0.6
2 / 0.6> x^2 - x
x^2 - x > 10/3
3x^2 - 3x - 10 > 0
x > [3+- (3^2-4*3*(-10))^(1/2)] / 2*3
x > (3 + (9+120)^(1/2)) /6
x > 2.39
Therefore x = 3
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)!]
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
It is 1 - 365Cn/365n. This is greater than 0.5 for n greater than or equal to 23.
Birthdays are not distributed uniformly over a year but if, for the sake of probability games you assume that they are, then ignoring leap years, the probability is 0.5687. Including leap years, it is slightly lower.
10^%
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.
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%
1-365/[(365-6)*365^6] = 1 Is this O.K ?
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)!]
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
It is 1 - 365Cn/365n. This is greater than 0.5 for n greater than or equal to 23.
Birth months are not uniformly distributed across the year. However, if yo assume that they are, the probability is 0.9536 (approx).
Birthdays are not distributed uniformly over a year but if, for the sake of probability games you assume that they are, then ignoring leap years, the probability is 0.5687. Including leap years, it is slightly lower.
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%
10^%
Using the hypergeometric distribution, the answer is 2114/3003 = 0.7040
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.