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?
This is a little tricky to do exactly analytically.
The barrels add up to 76 liters, so we are 6 liters short. Instead of distributing the 70 liters, let's distribute the 6 liter shortage. The complication is that all 6 liters of shortage cannot be distributed to any of the 5 liter barrels, but even if they were 6 liters this would be extremely low probability. Let's ignore that for now.
Then, the probability that the 1st liter of shortage goes to V is 2/12 = 1/6. The probability that it does not go to V is 5/6.
The probability that exactly 1 liter of shortage goes to V = 6*(1/6)*((5/6)^5) = (5/6)^5 = 3125/7776 ~ .4019. The same probability as rolling 6 dice and having exactly one show a 1.
Now, let's adjust for the impossibility of a 5 liter barrel being 6 liters short. The probability of trying to put 6 liters in one specific 5 liter barrel = (1/12)^6 = 1/2,985,984. There are 4 such barrels, so multiply by 4 to get 1/746,496 ~ .00000134. Subtracting this from the numerator and the demoninator gives us (.4019 - .00000134)/(1-.00000134) ~ .4019. In other words, it does not make a significant difference.
The exact fraction, I think, is (3125/7776 - 1/2985984)/(1-1/2985984) = 1199999/2985983 ~ .4019
Of course, I could be wrong