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Thousands of cricketers share the same birthday as another. If you were to specify one cricketer, it would be simple task to find several that share their birthday.
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How many people have the same birthday as you?
If there are 300,000,000 people in the US, then on average you share your birthday with over 82,000 people. Of course, if your birthday is 2/29, you would share it with about 1/4th of that number.
What If 2 folks share the same species what are all the levels they will share?
What If 2 folks share the same species what are all the levels they will share?
What is the probability that any two people in a random group of 30 have the same birthday that is the same month and date?
It is approaching unity that two in a random group of 30 people would have the same birthday. (Month and date)------------------------------------------------------------------------------------------------P(2 people share bd in 30) ≈ 38.0%Using the following expression* that doesn't consider February 29 of a leap year butgives good estimate;P(2 people and only 2 people share a birthday, month and date, in a group of n people) = nC2 (1/365) Π1n-1(1 - (i-1)/365)nCr = n!/(r!(n-r)!)for n = 30, r = 2,30C2 = 435P(2 people share bd in 30) = 435(1/365)(1)(1-1/365)(1-2/365)∙∙∙∙∙∙(1-27/375)(1-28/365) = 0.380215577...P(2 people share bd in 30) ≈ 38.0%*[For an expression that considers leap year see question "What is the probability thatin a room of 8 people 2 have the same birthday ?"]
Which president's birthday was on November 2nd 1865?
Warren Harding and James Polk were born on November 2. They are the only presidents who share a birthday.
What is the probability that two people selected from 30 have the same birthday?
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?"
How are the atoms of the elements in group 2 the same?
They share the same number of electrons (=2) in the valence shell: they have the same oxidation state of +2
Which 2 countries have the same countinent?
There are many countries which share the same continent. For example: Mexico and America share the same continent, but are two seperate countries.
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 2 continent share the same land?
Europe and Asia
What 2 continents share the same landmass?
Europe and Asia
What is the probabililty of at least 2 people same birthday from a group of 12 people?
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%).
If 50 people are chosen at random what is the probability that at least two of them have their birthday on the same day?
There being 365 days in a year and 50 being less than 365 therefore 2 even far less than that the chances are virtually 0. ______________________ Assume that all 366 days (including leap day) are equally likely to be a person's birthday. The probability that none of them share a birthday is 1*P(second person selected doesn't share a birthday with first person selected)*P(third person selected doesn't share a birthday with first or second person selected)*...*P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected). P(second person selected doesn't share a birthday with first person selected) = 365/366 P(third person selected doesn't share a birthday with first or second person selected)=364/366 . . . P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected)=317/366 P(none share a birthday)=(365/366)*(364/366)*...*(317/366), which is approximately .0299. P(at least two share a birthday) = 1-(365/366)*(364/366)*...*(317/366)=1-.0299=.9701 = 97.01%.
