(I) Conditional ProbabilityBayes' TheoremBayes' Theorem (Extended Form)(II) Independence(III) CombinatoricsFactorial NotationPermutationsCombinationsMath 370X - Lecture 2Conditional Probability, Independence, CombinatoricsSaumil PadhyaSeptember 12, 2016(I) Conditional ProbabilityP (B|A) =P (B ∩ A)P (A)→ P(B ∩ A) = P (B|A) · P (A)P (B) = P (B|A) · P (A) + P (B|Ac) · P (Ac)Example 1:Urn I contains 2 white and black balls and Urn II contains 3 white and 2 black balls.An Urn is chosen at random, and a ball is randomly selected from that Urn. Find the probabilitythat the ball chosen is white.If an exam question states that something is to be chosen at random and there is noindication otherwise, you should assume that each item has the same probability ofbeing chosen.Bayes’ TheoremP (A|B) =P (A ∩ B)P (B)=P (B|A) · P (A)P (B|A) · P (A) + P (B|Ac) · P (Ac)Example 2:Urn I contains 2 white and 2 black balls and Urn II contains 3 white and 2 blackballs. One ball is chosen at random from Urn I and transferred to Urn II, and then a ball ischosen at random from Urn II. The ball chosen from Urn II is observed to be white. Find theprobability that the ball transferred from Urn I to Urn II was a white.Bayes’ Theorem (Extended Form)P (Aj|B) =P (Aj∩ B)P (B)=P (B|Aj) · P (Aj)nPi=1P (B|Ai) · P (Ai)(II) IndependenceEvents A and B are independent ifP (A ∩ B) = P (A) · P (B)→ P (A|B) = P (A); P (B|A) = P (B)1Example 3:A blood test indicates the presence of a particular disease 95% of the time whenthe disease is actually present. The same test indicates the presence of the disease 0.5% of thetime when the disease is not present. One percent of the population actually has the disease.Calculate the probability that a person has the disease given that the test indicates the presenceof the disease.(III) CombinatoricsFactorial Notationn! = n(n − 1)(n − 2) · · · 2 · 10! = 1PermutationsGiven n distinct objects, the number of different ways in which the objects may beordered(orpermuted)is n!. For eg: the set of 3 letters {a,b,c} can be ordered in the following 3! = 6 ways:abc, acb, bac, bca, cab, cbaLet’s say we want to choose k ordered subsetswithout replacementout of n objects. The number of ways(permutations) you can achieve this isn!(n − k)!= n · (n − 1) · · · (n − k + 1)which is symbolically represented bynPk/Pn,k/P (n, k )For eg: The number of ways in which we can have a subset of 2 letter out of the set of 3 letters {a,b,c} is3!(3 − 2)!= 6ways → ab, ac, ba, bc, ca, cbCombinationsGiven n distinct objects, the number of ways of choosing a subset of size k≤nwithout replacementandwithout regard to the order isn!k! · (n − k)!which is symbolically represented by(nk)/nCk/Cn,k/C(n, k )2Number of Samples of r out of n objectsAttribute With Replacement Without ReplacementOrdered nrn!(n − r)!Unordered(n − 1 + r)!(n − 1)!r!n!r!(n − r)!Example 4:A and B draw coins in turn without replacement from a bag containing 3 dimesand 4 nickels. A draws first. It is known that A drew the first dime. Find the probability that Adrew it on the first
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