probability?

Question

From a population of people with unrelated birthdays,30 people are selected at random.WHat is the probability that exactly 4 people of this group have the same birthday and that all the others have different birthdays(exactly 27 birthdays altogether)?Assume that the birthrates are constant throughout the first 365 days of a year but that on the 366 th day it is one-fourth that of the other says

Answer 1

Maybe you can do this using combinations of 30 taken 4 at a time, but I don't know how to set that up without studying some stuff I haven't looked at in a long time.

There is another way, though. Since you want to consider "leap-year babies," there are 1461 days in a 4-year period. The chance of any given date, then, is 4 in 1461, or 1 in 1461 for 29 February.

Start with 1 January. The probability that any given person was born on 1 January is 4/1461, and the probabiity that he was born on that date is 1457/1461.

The probability that exactly 4 of 30 were born on 1 January is the combinations of 30 taken 4 at a time times (4/1461)^4, plus the combinations of 30 taken 26 at a time times (1457/1461)^26.

You need to repeat that for all 365 days, i.e., multiply by 365 since it's "or", then add in the probability that exactly 4 were born on 29 February, using the same technique as above.

This will give you the probability that exactly 4 of the 30 were born on the same day. Now you need to see if the remaining 26 were all born on different days. There are probably different approaches to this, but here's one.

Start with the knowledge that exactly 4 of the 30 were born on one day -- either 29 February or another day. Then, given that knowledge, you want to know the probabillity that on any other day, not more than one was born.

Ignoring the 29 February special case for the moment, consider the first of January, and assume that's not the date when four were born. Then the probability that any individual was born on 1 January is 4/1457, and the probability that an individual was born on that date is 1453/1457. Therefore the probability that zero or one were born on that date is (1453/1457)^26 + (1/1457)(1453/1457)^25.

You need to repeat (i.e., multiply) this by all remaining dates (364 of them), and also take care of the 29 February special case using these methods.

I suspect this is a fair amount of work but you ought to get an answer. There should be some patterns in the numbers that allow you to simplify quite a bit. And as I mentioned above, there may well be a more direct approach using combinations.

Answer 2

There is an easier way to take care of this (when distribution is uniform throughout the year):
the total amount of possible birthday combinations is 365^30 (^=to the power of)
the number of ways that at least 4 people are born the same day is 4 out of 30 = 30!/(30-4)!*4! (to choose which people...) times 365 (to choose the day)
substract the prob that 5 at least are born the same day and you get the answer.

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This very problem is trickier, since you have to take care of the 29th of February. Basically you need to differentiate between the case 29th Feb./ not 29th Feb. and do the same thing.