statisticalrethinkin..

56 2. SMALL WORLDS AND LARGE WORLDS2.6.12. A Tale of Two Pandas. Suppose there are two species of panda bear. Both are equallycommon in the wild and live in the same places. ey look exactly alike and eat the samefood, and there is yet no genetic assay capable of telling them apart. ey differ however intheir family sizes. Species A gives birth to twins 10% of the time, otherwise birthing a singleinfant. Species B births twins 20% of the time, otherwise birthing singleton infants. Assumethese numbers are known with certainty, from many years of field research.Now suppose you are managing a captive panda breeding program. You have a new femalepanda of unknown species, and she has just given birth to twins. What is the probabilitythat her next birth will also be twins?Solution. To solve this problem, realize first that it is asking for a conditional probability: Pr(twins 2 |twins 1 ),the probability the second birth is twins, conditional on the first birth being twins. Remember thedefinition of conditional probability:Pr(twins 2 |twins 1 ) = Pr(twins 1, twins 2 )Pr(twins)So our job is to define Pr(twins 1 , twins 2 ), the joint probability that both births are twins, and Pr(twins),the unconditioned probability of twins.Pr(twins) is easier, so let’s do that one first. e “unconditioned” probability just means that wehave to average over the possibilities. In this case, that means the species have to averaged over. eproblem implies that both species are equally common, so there’s a half chance that any given pandais of either species. is gives us:Pr(twins) = 1 2} (0.1) + 1 2{{ }(0.2) .} {{ }Species A Species BA little arithmetic tells us that Pr(twins) = 0.15.Now for Pr(twins 1 , twins 2 ). e probability that a female from species A has two sets of twins is0.1 × 0.1 = 0.01. e corresponding probability for species B is 0.2 × 0.2 = 0.04. Averaging overspecies identity:Pr(twins 1 , twins 2 ) = 1 2 (0.01) + 1 2(0.04) = 0.025.Finally, we combine these probabilities to get the answer:Pr(twins 2 |twins 1 ) = 0.0250.15 = 25150 = 1 6 ≈ 0.17.Note that this is higher than Pr(twins). is is because the first set of twins provides some informationabout which species we have, and this information was automatically used in the calculation.2.6.13. Name that panda. Recall all the facts from the problem above. Now compute theprobability that the panda we have is from species A, assuming we have observed only thefirst birth and that it was twins.Solution. Our target now is Pr(A|twins 1 ), the posterior probability that the panda is species A, giventhat we observed twins. Bayes’ theorem tells us:Pr(A|twins 1 ) = Pr(twins 1|A) Pr(A).Pr(twins)We calculated Pr(twins) = 0.15 in the previous problem, and we were given Pr(twins|A) = 0.1 aswell. e only stumbling block may be Pr(A), the prior probability of species A. is was also given,