PRACTICE WITH CLIL

10 ESERCIZI Eventi e probabilità 79 Uno studente si presenta poco preparato a un esame e valuta di poter svolgere da solo il compito con probabilità del 40% e di poter copiare con probabilità del 60%. Data la sua scarsa preparazione, la probabilità di fare giusto il compito è del 35% e, se copia, ha invece probabilità del 30% di copiare da un compagno che sa fare il compito. Nell ipotesi che il compito sia stato svolto correttamente, calcola la probabilità che lo studente [43,75%] l abbia fatto da solo. 80 Un test diagnostico è in fase sperimentale e attualmente fornisce esito positivo nel 90% dei casi in cui una certa disfunzione è effettivamente presente e anche nell 1% dei casi in cui la disfunzione non è presente. Da ricerche collaterali, si sa che la percentuale di soggetti che hanno tale disfunzione è il 3% della popolazione diagnosticata. Calcola la probabilità che la disfunzione sia comunque presente anche nel caso in cui il test [0,31%] abbia dato esito negativo. PRACTICE WITH CLIL Probability and genetics The black trait (which gene has been capitalized) is dominant, while the white trait is recessive. This means that if the pair of alleles (called genotype) has even a single black gene, the physical appearance of the animal, with respect to this trait (called phenotype), is dominated by this gene: the animal is black haired. The phenotype is instead white only if the genotype is bb (two white genes). Let us summarise the situation: genotype NN phenotype: black genotype Nb phenotype: black genotype bb phenotype: white When two guinea pigs mate, each offspring randomly receives one allele from the male parent and one allele from the female parent. In front of a white guinea pig it is immediate to establish its genotype: it can only be of the bb type. Instead, with a black guinea pig, there is some uncertainty: two are the possible genotypes, NN or Nb. Given a black male guinea pig, we want to sketch a series of experiments to establish its genotype with a 95% confidence level. To avoid a multiplication of hypotheses, the male is mated with a white female, which certainly has a bb genotype. Therefore, we have the following probability tree (a.), which second string of possibilities represents the phenotypes of the offspring born from mating. 1 2 male NN female bb 1 a. Nb = black 1 2 1 2 male NN female bb male Nb female bb 1 1 2 2 Nb = black bb = white 1 b. k babies born all black 1 2 male Nb female bb 1 2k k babies born all black 1 1k 2 others In this case, as soon as a white baby is born the problem is solved: the male pup is definitely an Nb genotype. However, if a white baby is not born, how many black pups will have to be born so that we can be 95% confident that the male pup has the NN genotype? 1 Suppose k babies were born, all black, one after the others. The probability that this distribution occurs is __k . We 2 can therefore rewrite the probability tree (b.) as if k black offsprings were born at the same time (for calculating the probabilities, it is indifferent whether the information is given one pup at a time, or all together): 1 __ 1 ___ 2k 2 485

Il Maraschini-Palma - volume 4
Il Maraschini-Palma - volume 4