Student: Adrien de Chilly 1. Há quantos anos existem as seguintes áreas do conhecimento: Física Química Biologia Matemática Estatística Ciência da Computação Answer To begin, we must point out the evolution of philosophy and its splitting into two branches that we shall call natural philosophy and spiritual philosophy. The former one is what has evolved into science, while the latter can be called moral philosophy, and is what we commonly refer to when talking today about philosophy. Mathematics started with demonstration, and therefore Egyptians and Mayas cannot be counted as founders of Mathematics. We shall take Pythagoras for example, dating back to about 2500 years. Physics is much more recent, because we must consider the point where the knowledge people had is still right. No Aristotle then. Galileo, Newton or Kepler can be thought of as precursors to what we now know of Physics, having it date back to around 400 years. Biology date back to the description of plants, their naming, around 500 years ago. Did Chemistry start with Lavoisier or more recently Mendeleev? We shall go with Lavoisier, who experimented and came up with what is now Lavoisier's law: "Rien ne se perd, rien ne se crée, tout se transforme". People had knowledge before, but they did not try to explain what was happening, therefore alchemy cannot be considered as chemistry. Chemistry is 200-250 years old. Statistics can be traced back to Bernoulli, if we regroup Probability with Statistics, making it 400 years. If we count only modern statistics, we will go back to 100 years at most. Computing can be traced back to 50 or 100 years, or even further away if we take into account Charles Babbage. Can we go back as far as punchcards in the 18th century ? Or to Alan Turing ? We shall keep both 50 and 100 years, noticing it is still very close when compared to all the other sciences afore mentioned. To summarize, we have: Physics 400 years Chemistry 200-250 years Biology 500 years Matemathics 2500 years Statistics: Probability, 400 years; Modern statistics, 100 years Computer Science 50-100 years 2. Um jogo de Mega-Sena consiste em marcar seis números dos sessenta existentes no volante. No sorteio são escolhidos também seis números. Qual a probabilidade de se acertar? We are looking to pick out 6 precise balls out of 60, in no particular order. This looks very much as a combination of 60 and 6. This result is 60!/(54!*6!). This is of course having supposed that all the balls are the same, and that the odds of picking each ball are equal, therefore a probability of 1/60. Here is an easier explanation: - 1st ball: we can pick any of the 6 among the 60. The probability to pick one of the 6 is therefore 6*1/60, or 6/60. - 2nd ball: we assume that we have picked one of the intended balls: There are 59 balls left, and we must pick one of the 5. The probability of picking one of the 5 balls out of 59 is 5/59. And so on for the next balls, assuming we are very lucky and pick one of the original 6 each time ... - 3d ball: 4/58 - 4th ball: 3/57 - 5th ball: 2/56 - 6th ball: 1/55 The winning probability is (6/60)*(5/59)*(4/58)*(3/57)*(2/56)*(1/55), which is indeed 60!/(54!*6!). Of course, the order is not important, or the winning chances would be too small. In scientific notation, the result is: 1.997449*10^-8. About 20 chances in a billion. 3. Usando o resultado do exercício anterior, suponha que alguém jogue 1 milhão de vezes na Mega-Sena. Qual é o número esperado de vitórias? To find out the number of wins out of a million games, we must calculate the expected value of a random variable representing the number of wins that many games. To begin with, consider variable X : Omega -> R. Omega = {w,l}, with w for win and l for lost. The probability to win is P, and to lose 1-P. X : w -> 1 l -> 0 Variable X represents the number of wins in a single game. Its expected value E(X) is the sum of xi P(X=xi) which is 1*P+0*(1-P)=P. Variable Y, the number of wins in 1,000,000 games, is the sum of Xi, with i going from 1 to 1,000,000, where each Xi has the same distribution as X. Therefore, E(Y) is the sum of the expected values, therefore 1,000,000*P, or about 0,02.