Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
The birthday paradox, a question?
Suppose n people are queued, where n is a positive integer less than 366. Each person is asked in turn to write down their birthday (month and day). We assume no one in the queue was born in a leap year, and that every possible birthday is equally likely.
(e) Suppose n = 30. What is the probability that two of the 30 people in the queue have the same birthday? (This counter intuitive result is called the Birthday Paradox).
(f) Suppose n = 30 and the first person has already recorded their birthday. What is the probability that one of the 29 people still in the queue has the same birthday as the first? Why do your answers to parts e and f compare the way they do?
Attempt:
For part (e) I tried 1 - (1-1/365)(1-2/365)*...*(1-30/365) = 0.706
For part (f) I have no clue....
1 Answer
- ?Lv 77 years agoFavorite Answer
(E) Your answer to this part is correct.
(F) Suppose person #1, say, has some birthday. The probability that person #2 does NOT the same birthday as person #1 is 364/365. In a similar fashion, person #n with n ≥ 2 do not have the same birthday as person #1 with probability 364/365. Hence, the required probability is:
1 - P(persons #2-#29 do not have the same birthday as #1) = 1 - (364/365)^29.
As to why these differ significantly, notice that part (F) only compares the birthdays of 29 people to one specific person, whereas part (E) compares the birthdays of all 30 people to each other. As a result, the probability is much greater, as a group of 30 people satisfies the property in part (E) if ANY person in the group matches a birthday out of the other 29, versus the 29 people having to match 1 birthday.