BIOL 570 1nd Edition Lecture 6Outline of Last Lecture I. Review of chapter 4II. ProbabilityIII. Definitions: random trial, sample space, event, probability, mutual exclusive eventsIV. Probability Rules 1-5 Outline of Current Lecture I. IndependenceII. Multiplication ruleIII. Conditional probabilityIV. Bayes’ ruleV. Law of Total ProbabilityCurrent LectureIndependent- if knowing that one event occurs tells you nothing about the probability of a second event, then the two events are independent of each otherPr (A|B) = Pr (A|not B) = Pr (A)*P (A|B): probability of event A given that event B occurredRule #6If A and B are independent events then: Pr (A and B) = Pr (A) Pr (B)What is the probability that the next child born in the US will be female and will get an odd SSN?Pr (F) = ½Pr (odd SSN) = ½ Pr (F and odd SSN) = ½ * ½ = ¼ These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.The purple petal allele (P) is dominant to white (p). Round pea allele (R) is dominant to wrinkled (r). Petal trait is inherited independently of round/inherited trait. What is the probability that the offspring of a PpRr x PpRr cross will have white flowers and round peas?P (phenotype = round) = P (genotype = RR or Rr)P (geno = RR) = P (mom give R and dad gives R) P (m.g.R.) x P (d.g.R.) ½ * ½ = ¼ Approximately 16.5% of males will develop prostate cancer. Assuming 50:50 gender ratio at birth, what is the probability that a randomly chosen infant will be male and will have prostate cancer?The general multiplication rule (works whether or not A and B are independent). Rule #7P (A and B) = P (A) P (B|A) = P (B) P (A|B) P (M and P.C) = P (M) P (P.C|M)= ½ (.165) = .0825What is the probability of a woman becoming pregnant at least once over a ten-year time span?(no contraception) For a single year:Pr (pregnancy | no contraception) = 0.85P: pregnantN: not pregnantP (not P) = P (N) = 1 – P (in one yr) = 1 – 0.85 = 0.15P (not N1 N2 N3 N4 N5 N6 N7 N8 N9 N10) = P (at least one P) = 1- P (N1 N2 N3 N4 N5 N6 N7 N8 N9 N10)*P (N1) is independent of P (N2)P (N10) = P (N1) P (N2) P(N3)… P(N10) → P(N)10Pr (P in 1 | None) = 0.85Pr (not P in 1 | None) = 0.15Pr (not P in 10 | None) = Pr (no p in 1 | None)10Pr (P in 10 | None) = 1 – Pr (not P in 10 | None)Pr (at least 1 P in 10 | None) = 1 – 5.8x10-09A disease is found in 8% of the population. The test has a false positive rate of 10% and a false negative rate of 5%. If someone has a positive result, what is the chance that they have the disease?D: diseaseH: healthyP (D T=+ ) = ? = [P (+ | D) P(D)] / P(+)P (D) = 0.08P (+) = ?P (+|H) = 0.1P (- | D)= 0.05P (+ | D) = 1- P (- | D) = 0.95P (H) = 1- P (D) = 1 – 0.08 = 0.92Baye’s RuleP (B|A) = [P (B) P (A|B)]/ P(A)Law of Total ProbabilityP (A) = Ʃ P (Ei) P (A|Ei) if S= {Ei, E2……Ek}, if the sample space of event EP (+) = P (D) P (+|D) + P(H) P(+|H)= (0.08)(0.95) +