The probability that 12 randomly selected people from a group of 12 men and 18 women will all be women is (18 in 30) times (17 in 29) times (16 in 28) times (15 in 27) times (14 in 26) times (13 in 25) times (12 in 24) times (11 in 23) times (10 in 22) times (9 in 21) times (8 in 20) times (7 in 19), which is equal to 8,892,185,702,400 in 2,180,547,008,640,000, which is equal to 68 in 16,675,which is equal to 0.00407796.
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)!]
no
Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082
the larger the group, the more likely the statistical probability of loss will be equal
To select the first birthday, the probability is 1/30. Having gotten that, the conditional probability that the next birthday would be the same is (1/30)x(1/29) and that is 1/870----------------------------------------------------------------------------------------------I believe the question has to be rephrased to "What is the probability that two peoplein a group of 30 people share the same birthday?". Because in the way the questionis actually stated, "the probability that two persons selected randomly from a group of 30 have the same birthday", the event that "those two people would share theirbirthday" is independent of the size of the population they were selected from.In the case the actual question be "What is the probability that two people in a groupof 30 share the same birthday?, is given by the following expression that neglectsFebruary 29 (of the leap), but gives very good approximation to the expression thatconsiders February 29 and is a simpler one. [It has to be mentioned that the analysisleading to this expression considers birthdays a "random variable" where chances fora persons birthday are the same for any day of the year]:P(2 share bd out of n) = nC2 (1/365) Π1n-1[1-(i-1)/365]for n = 30, P(2 share bd out of 30) = 30C2 (1/365) Π1 29 [1-(i-1)/365] = 435∙(1/365)∙[1-1/365]∙[1-2/365]∙[1-3/365]∙ ∙∙∙ ∙[1-28/365] = 0.380215577... ≈ 0.380 ≈ 38.0%For the construction of the expression to calculate the probability of any two people sharing a birthday in a group of n people considering Feb 29 of the leap year see thequestion "What is the probability that in a room of 8 people 2 have the same birthday?"
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.
a select group
The answer depends on what is the probability of WHAT!
christians
Only a select group of people.
Caucus
no
Birth months are not uniformly distributed across the year. However, if yo assume that they are, the probability is 0.9536 (approx).
The Electoral College
The probability of at least 2 people in a group of npeople sharing a common birthday can be expressed more easily (mathematically) as 1 minus the probability that nobody in the group shares a birthday. Consider two people. The probability that they don't have a common birthday is 365/365 x 364/365. So the probability that they do share a birthday is 1-(365/365 x 364/365) = 1-365x364/3652 Now consider 3 people. The probability that at least 2 share a common birthday is 1-365x364x363/3653 And so on so that the probability that at least 2 people in a group of n people having the same birthday = 1-(365x363x363x...x365-n+1)/365n = 1-365!/[ (365-n)! x 365n ]In the case of 12 people this equates to 0.16702 (or 16.7%).
3/10 or 0.3