Part A. Imagine that two parents have two children, one who produces 4 grandchil
ID: 3070929 • Letter: P
Question
Part A. Imagine that two parents have two children, one who produces 4 grandchildren and the other who produce 2 grandchildren. What is the chance that there is a progression parent-child-grandchild sharing birthdays? To simplify things, you can assume the possibility of this happening more than one way is realtively negligible, including the possiblity of muliple births (such as twins.)
Part B. What is the chance that three people in the family of ten share a birthday? (Again, ignore mulitple scenariois in the same family.)
Part C. Continuing to assume as in parts A and B, and suppose three people in the family of 10 share a birthday. What is the chance they are in a parent-child-grandchild progression?
Part D. Obviously families come in different sizes. But suppose there is approximately a $10^{-4}$ chance of a grandparent-parent-child sequence sharing a birthday in a family and families are independent. What is the chance that this occurs in at least one family for acity of 30,000 families?
Explanation / Answer
For any given 3 people for which we can reasonably expect that their birthdays are relatively effectively random, it will be about 1 in (365*365). However, you have not told us how many different subsets of 3 can be selected from you family. To a first approximation, you can just multiply by that number.
There is no sample size given in the question. Three people share their bday out of how many people??
Lets say if there are 87 people, the probability of at least three people sharing a birthday is very close to 0.50 (making the standard assumption of an uniform distribution over 365days).
On the face of it, for three particular people independently to have the same birthday, the probability is about 1/(365)(365) (rather more because they are not independent if families are planned).
What's the probability that, in a family of ten, exactly three people have the same birthday?
For a family of ten: (365C8 * 10! / 3!) / 365^10 =
((7232294429652435 * 3628800)/6)/4.1969002243198805e+25) =
0.00010422195 which is 1 in every 10,000