No Natal do ano passado, durante uma longa noite de entrega de presentes, o Papai Noel chegou, muito cansado, a uma rua com 7 casas. Então, ele resolveu atirar todos os 7 presentes aleatoriamente. Dado que cada casa recebe um presente, e que os presentes são todos diferentes entre si, qual é a probabilidade de o Papai Noel preguiçoso atirar pelo menos dois presentes nas casas corretas?
(In Christmas last year, during a long night of gift delivering, Santa Claus arrived, very tired, at a street with 7 houses. Then, he decided to throw all the 7 gifts randomly. Given that each house receives one gift, and that the gifts are all different from each other, what is the probability that lazy Santa Claus throws at least 2 gifts to the correct houses?)
PCFilho
Since Santa is distributing the presents randomly, and that there are 7 presents distributed this way, that means that for each present distributed the probability that it's the correct present for that house is 1/7. The probability of all the gifts being distributed incorrectly is (6/7)^7. The probability of only one house receiving the correct present is 7*(1/7)*(6/7)^6. So, the probability of at least 2 houses receiving the correct present is 1-(6/7)^7-(6/7)^6, or approximately 26.4%.
ResponderExcluirI expect that there were five children who had a very disappointing Christmas that year. XD
ExcluirHopefully the neighbors are good friends and will exchange the wrong gifts. HA HA HA!
ExcluirWell done, Jake!! I calculated it in a different way, counting the possibilities, but fortunately we got the same result. :)
ResponderExcluirThe total number of possibilities of gift distributions is 7! = 5040.
The number of possibilities in which all gifts were thrown to the wrong houses is !7 = 1854.
The number of possibilities in which only one gift was thrown to the correct house is 7*(!6) = 7*265 = 1855.
The possibilities in which at least two gifts were thrown to the correct houses is, therefore: 5040 - 1854 - 1855 = 1331.
The probability of throwing at least two gifts to the correct houses is, therefore, 1331/5040. Or approximately 26.4%. :)
Note: by !n, I mean the number of derangements of n. A derangement is a permutation of the elements of a set, such that no element appears in its original position.
ExcluirThere are different ways of counting the number of derangements. Starting with n = 2, the numbers of derangements of n are:
n = 2: 1
n = 3: 2
n = 4: 9
n = 5: 44
n = 6: 265
n = 7: 1854
n = 8: 14833
n = 9: 133496
n = 10: 1334961
n = 11: 14684570
n = 12: 176214841
n = 13: 2290792932
And so on. In the On-line Encyclopedia of Integer Sequences, it is the sequence A000166.
PS: I have already posted a problem of counting derangements here in the blog: Amigo oculto.
ExcluirHere is a way to calculate the number of derangements, recursively:
ExcluirD[N]=(N-1)*(D[N-1] + D[N-2])
D[1] = 0
D[2] = 1
D[3] = 2*(D[2] + D[1]) = 2*(1 + 0) = 2
D[4] = 3*(D[3] + D[2]) = 3*(2 + 1) = 9
D[5] = 4*(D[4] + D[3]) = 4*(9 + 2) = 44
D[6] = 5*(D[5] + D[4]) = 5*(44 + 9) = 265
D[7] = 6*(D[6] + D[5]) = 6*(265 + 44) = 1854
And so on...