Birth Date
Suppose there are 40 students in a class, and suppose they are all born in the same year. What are the chances that two students have same birth date ?
Labels: logic
posted by Rajesh Lal at
2:14 AM
Friday, January 23, 2009
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23 Comments:
I'm not positive but I think the problem is a combination of 40C2.
40!/(38!2!)=780
Hello,
Well, the question states what are the chances of two having the same birth date...
So as there is no reason why two couldn't have the same birth date I would say the answer is:
100%
Windy
very low.
from my experience there is always at least one person with the same as another
There is a very high chance that they will two will have the same birth dates.
With just two students the chances are 364/365. The student has only one other birth date to match with.
Mathematically with only 23 students there is approximately a 50% chance that two will have the same birthday. There are a lot of possible matches – the first person has 22 others to match with, the second person 21 others to match with, as well as with the first person, and so on – in the end there are over 250 combinations of possible matches.
It works out that with 40 people the chances are very nearly 90% certainty of a match.
Of course to be 100% certain you would need 366 students (or 367 if their birth year was a leap year) to be absolutely certain to have a match.
thats not strictly true Questeruk, its still possible for all 366 students to be born on the same day..
this question has to be solved using bayes theorem.. which i really dont want to do right now!
The problem is slightly ambiguous:
1. it is not clear whether more than two students having the same birth date is included, and
2. it is not stated whether the year of their birth is a leap year.
Let's assume that the question is '...two or more students have the same birthday' and the year is non-leap (i.e. 365 days).
Let X = the chances of 2 or more students having the same birthday.
Then 1-X is the probability of all birth days being distinct.
In order for N students to have distinct birthdays, the first N-1 must have distinct birth days and the Nth must have a different birth day. That is the Nth student must hit a chance of (365-(N-1))/365, since N-1 days are taken up by the first N-1 students.
So,
1-X =
(365 - (1-1))/365 *
(365 - (2-1))/365 *
(365 - (3-1))/365 *
...
(365 - (40-1))/365
=
1 *
364/365 *
363/365 *
...
326/365
A quick calculation with your favorite programming language (or a slower one with a calculator) shows:
1-X = 0.108768..., so
X = 0.891231... or almost 90%.
It is rather unintuitive, that it takes a number of people, that is relatively small compared with the number of days in a year, to have a high chance of this happening. So this is known as the birthday paradox.
take one student as a reference student who was born on day y. the odds of another student being born on that date are 1/365. the odds of another student being born that day is also 1/365. the odds that both the initial student shares a date with either of these two student are 2/365 (surprisingly enough, 1+1=2). Since there are in total 39 other students, the odds of him sharing a birthdate with any of the other 39 students are 39/365
jp
very low chance
I have heard this one.The answer is 97%.
Jp, 39/365 is the chances of ONE specific student having a birth date the same as another of the 39.
Surge did the maths that show there is a 90% chance of ANY two students having the same birth date, which is the question posed.
Since there are 40 stucents and 365 days in a year, there would be a 40 in 365 chance that two of the students' birthdays were the same day.
Although, with a 4 in 1461 chance that any of the students were born on February 29th, the odds explode to 40 in 14,610 chance that any two of the student were both born on Leap Day.
Still, the overall chance that two were born on the same day in 40 in 365, unless there are twins enrolled in the school.
i think the answer would be 1/(365X40) equal to 1/14600 coz each of the student only have one birthday of the year and there are 40 student in the rooms. what do you all think.
you guys are putting too much thought into this, its a logic question
1/365 = 0.002739% chance of someone with an identical birthday. even if all 10 students have the same birthday each student will have the same odd's
There's lots of math on Wikipedia about this problem, (birthday paradox) and the answer for 40 people is almost 90%.
Some here have confused one person matching one of the other 39... that's not right. The question is any of the 40, matching any other of the 39. So if you have computed around 2% then you need to multiply that by 40.
There is a very high chance that they will two will have the same birth dates.
With just two students the chances are 364/365. The student has only one other birth date to match with.
Mathematically with only 23 students there is approximately a 50% chance that two will have the same birthday. There are a lot of possible matches – the first person has 22 others to match with, the second person 21 others to match with, as well as with the first person, and so on – in the end there are over 250 combinations of possible matches.
It works out that with 40 people the chances are very nearly 90% certainty of a match.
Of course to be 100% certain you would need 366 students (or 367 if their birth year was a leap year) to be absolutely certain to have a match.
I don't really know what I did, but I got 9,1%
Hey Anonymous of
January 25, 2009 5:37 PM
Do you think that maybe you just copied my answer further up this thread?
( posted January 23, 2009 10:19 AM under Questeruk) – it’s word perfect to my posting, including the mistype in the first paragraph!
Now what are the chances of that????
In computer hacking, this is a known issue, refer to:
http://en.wikipedia.org/wiki/Birthday_attack
The birthday paradox is well known. The calculations that get 90% are correct. It's actually a even higher than that if you surveyed actual classes, because the distribution of birthdays is not uniform (more birthdays July-October), and with all in same class, the odds of twins is significant. There's a 3% chance of any one kid being a twin, a remarkably high number in itself. This means in a class of 40, the odds of there being a twin in the class are very close to 50/50.
i was in that class and allison had my birthday too
and Mrs.Allexan got us confused
I think that its 2/40 now i mayby wrong but I am a guenis
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