STA 1300 Lecture Chapter 10

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Apr 3, 2024

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1 10. Independence and Conditional Probabilities 1. Relationship among Several Events: When the two events (A,B) cannot occur together, these events are disjoint, or mutually exclusive. Therefore, P (A and B) = 0. Suppose that you want to study the next two single births at a local hospital. You are counting girls, so two events of interest are A = first baby is a girl B = second baby is a girl. The events A and B are not disjoint (they are not mutually exclusive). They occur together whenever the next two single births at the hospital are girls. We can display all possible combinations of outcomes in a two-way table: Ex) Forests are complex, evolving ecosystems. For instance, pioneer tree species can be displaced by successional species better adapted to the changing environment. Ecologists mapped a large Canadian forest plot dominated by pioneer Douglas fir with an understory of the invading successional species western hemlock and western red cedar. The following two-way table records the distribution of all 2050 trees in the plot by species and by life stage. The distinction between live and sapling trees is made for live trees taller or shorter than 1.3 meters, respectively.
2 (ex) Characteristics of a Forest Dead Live Sapling Total Western Red Cedar (RC) 0.02 0.10 0.08 0.20 Douglas Fir (DF) 0.16 0.16 0.00 0.32 Western hemlock (WH) 0.23 0.21 0.04 0.48 Total 0.41 0.47 0.12 a. The probability that randomly selected tree is Western Red Cedar? P(RC) = P(RC and Dead) + P(RC and Live) + P(RC and Sapling) = 0.02 + 0.10 + 0.08 = 0.20 b. The probability that randomly selected tree is Western Hemlock? P(WH) = c. The probability that randomly selected tree is Sapling? P(Sapling) = 2. Conditional Probabilities: P(B|A) = 𝐏(? 𝐚𝐧𝐝 ?) 𝐏(?) or 𝐏(? 𝐚𝐧𝐝 ?) 𝐏(?) = P(A|B) Now we are told that the tree selected is a sapling. That is, it is one of the 12% of all trees that fall in the “Sapling” column of the table. The probability that a tree is a western hemlock, given the information that the tree is a sapling, is the proportion of western hemlocks in the “Sapling” column: P (WH | sapling) = P (WH and sapling) / P (sapling) = 0.04/0.12 = 0.3333 This is a conditional probability. You can read the bar ∣ as “given the information that.”
3 a. Find the probability that a randomly selected tree is sapling given that it is “ WH ”. P (sapling|WH) = P (sapling and WH) / P (WH) = 0.04/0.48 = 0.0833 b. Find the probability that a randomly selected tree is DF given that it is “Live”. P (DF |Live) = c. Find the probability that a randomly selected tree is DF given that it is “Dead”. d. Find the probability that a randomly selected tree is Live given that it is “WH”. e. Find the proba bility that a randomly selected tree is DF given that it is “Live”. f. Find the probability that a randomly selected tree is RC given that it is “Dead”. g. Find the probability that a randomly selected tree is Dead given that it is “RC”. 3. General Probability Rule (1) General Addition Rule 𝑷 ( A or B ) = 𝑷 ( A ) + 𝑷 ( B ) – 𝑷 ( A and B )
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4 (ex) Characteristics of a Forest Dead Live Sapling Total Western Red Cedar (RC) 0.02 0.10 0.08 0.20 Douglas Fir (DF) 0.16 0.16 0.00 0.32 Western hemlock (WH) 0.23 0.21 0.04 0.48 Total 0.41 0.47 0.12 a. Find the probability that a randomly selected tree is sapling or that it is “ WH ” . P (sapling or WH) = P (sapling) + P (WH) − P (sapling and WH) = 0.12 + 0.48 – 0.04 = 0.56 b. Find the probability that a randomly selected tree is DF or that it is “Live”. P (DF or Live) = c. Find the probability that a randomly selected tree is DF or that it is “Dead”. P (DF or Dead) = d. Find the probability that a randomly selected tree is Live or that it is “WH”. P (Live or WH) = Ex) Congenital sensorineural deafness is the most common form of deafness in dogs and is often associated with congenital pigmentation deficiencies. A study of hearing impairment in dogs examined thousands of Dalmatians for both hearing impairment and iris color. “Impaired” was defined as deafness in either one or both ears. Dogs with one or both blue irises (a trait due to low iris pigmentation) were labeled “blue.” The study found that 28% of the Dalmatians were hearing-impaired, 11% were blue-eyed, and 5% were hearing-impaired and blue-eyed. Choose a Dalmatian at random. What is the probability that the dog has blue irise(s) or has hearing impairment?
5 4. General Multiplication Rule: P(B|A) = 𝐏(? 𝐚𝐧𝐝 ?) 𝐏(?) P(A|B) = 𝐏(? 𝐚𝐧𝐝 ?) 𝐏(?) From this, P(A and B) = P(A) P(B|A) P(A and B) = P(B) P(A|B) In a 2012 Gallup survey of a random sample of American adults, 52% said they do not drink any soda on a typical day. Of those who said they do not drink soda on a typical day, 39% described themselves as overweight. Let’s consider this sample survey to be representative of the U.S. adult population. What percent of American adults drink no soda and are overweight? Use the general multiplication rule: P(no soda) = 0.52 P(overweight ∣ no soda) = 0.39 P(A) = 0.52 P(B ∣ A) = 0.39 P(no soda and overweight) = P(no soda)× P(overweight ∣ no soda) P(A and B) = P(A)× P(B ∣ A) =

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