Under French law, the Beaujolais Nouveau
( wine ) is released at 12:01 A.M.
on the third Thursday in November every year.
Last year, prior to the above date 11 wine amateurs deposited 12 small-size barrels at their merchant’s store (A,B,C,D,E,F,G - 7 liters each; S,T,U - 5 liters each and V two barrels: one 7 liters and one 5 liters).
When the wine arrived 70 liters were poured in the above barrels so that each barrel got an integer number of liters. The Customers were billed accordingly.
If every possible distribution of wine among the 12 barrels is equally likely, what is the possibility that V(ictor) had to pay for 11 liters of wine?
(In reply to One second thought ...
by Steve Herman)
OK, second try.
Start with 76 liters of wine, fill all the barrels, and them remove 6 liters.
Distribute the 6 liters (shortages) among 12 barrels. Imagine 6 objects and 11 separators, for a total of 17 things. There are a total of 17!/(6!*11!) = 12376 arrangements, each one corresponding to a different distribution of shortages to barrels. 4 of these are invalid, because they would take 6 liters out of a 5 liter barrel. 12,372 ways of doing this legitimately.
And how many of these result in removing exactly one from one of V's two barrels? It is equal to the number or ways of distributing 5 liters (shortages) among 10 barrels, times the number of ways of distributing 1 liter among two barrels. This is 14!/(5!*9!) * 2 = 4004. These are all valid, so
the final probability = 4004/12372 = 1001/3093 ~ .3236
Edited on October 22, 2014, 3:00 pm