9. Introducing probability The Practice of Statistics in the Life Sciences Third Edition © 2014 W. H. Freeman and Company

Objectives (PSLS Chapter 9) Introducing probability Randomness and probability Probability models Discrete vs. continuous sample spaces Probability rules Continuous random variables Risk and odds

Randomness and probability In a random event, outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. We define the probability of any outcome of a random phenomenon as the proportion of times the outcome would occur in a very long series of repetitions. The sex of a newborn is random P(Male) ≈ 0.512, the proportion of times a newborn is male in many, many births. Year 2002 2004 2006 2008 Males 2,058 2,105 2,184 2,173 Females 1,964 2,007 2,081 2,074 Proportion of males 0.5117 0.5118 0.5121

Probability models Probability models mathematically describe the outcome of random processes. They consist of two parts: 1) S = Sample Space: This is a list or description of all possible outcomes of a random process. An event is a subset of the sample space. 2) A probability assigned for each possible simple event in the sample space S. Probability model for the sex of a newborn S = {Male, Female} P(Male) = 0.512; P(Female) = 0.488 Year 2002 2004 2006 2008 Males 2,058 2,105 2,184 2,173 Females 1,964 2,007 2,081 2,074 Proportion of males 0.5117 0.5118 0.5121

Discrete vs. continuous sample spaces Discrete sample space: Discrete variables that can take on only certain values (a whole number or a descriptor). Blood types For a random person: S = {O+, O-, A+, A-, B+, B-, AB+, AB-} Continuous sample space: Cholesterol level For a random person: S = any reasonable positive value (or the interval zero to infinite) Continuous variables that can take on any one of an infinite number of possible values over an interval.

B B - BBB G … G G - BBG B - BGB G - BGG … S = { BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG } A. A couple wants three children. What are the possible sequences of boys (B) and girls (G)? B. What is the number of days last week that a randomly selected teen exercised for at least one hour? S = { 0, 1, 2, 3, 4, 5, 6, 7 } A and B are examples of discrete sample spaces. C is an example of a continuous sample space. C. A researcher designs a new maze for lab rats. What are the possible outcomes for the time to finish the maze (in minutes)? S = interval zero to infinity (all positive values)

Probability rules Probabilities range from 0 (no chance) to 1 (event has to happen): For any event A, 0 ≤ P(A) ≤ 1 The probability of the complete sample space S must equal 1: P( sample space ) = 1 The probability that an event A does not occur (not A) equals 1 minus the probability that is does occur: P( not A ) = 1 – P(A)

The 2011 National Youth Risk Behavior Survey provides insight on the physical activity of U.S. high school students. Physical activity was defined as any activity that increases heart rate. Here is the probability model obtained by asking, “During the past 7 days, on how many days were you physically active for a total of at least 60 minutes per day?” What is the probability that a randomly selected U.S. high school student exercised at least 1 day in the past 7 days? A) 0.08 B) 0.23 C) 0.77 D) 0.85 E) 0.92

Disjoint events Two events are disjoint, or mutually exclusive, if they can never happen together (have no outcome in common). Events A and B are disjoint. Events A and B are NOT disjoint. “male” and “pregnant”  disjoint events “male” and “Caucasian”  not disjoint events

Addition rules Addition rule for disjoint events: When two events A and B are disjoint: P(A or B) = P(A) + P(B) General addition rule for any two events A and B: P(A or B) = P(A) + P(B) – P(A and B)

Probability that a random person is type O+ P(O+) = 0.38 P(all blood types) = .38 + .07 + .34 + .06 + .09 + .02 + .03 + .01 = 1 Probability that a random person is not type A+ P(not A+) = 1 – P(A+) = 1 – 0.34 = 0. 66 Probability that a random person is “blood group O” P(O) = P(O+ or O-) = P(O+) + P(O-) = .38 + .07 = .45 Probability that a random person is “rhesus neg” P(O- or A- or B- or AB-) = .07 + .02 + .06 + .01 = 0.16 Probability that a random person is either “blood group O” or “rhesus neg” P(O or -) = P(O) + P(neg) – P(“O-”) = .45 + .16 – .07 = .54

HI = Dalmatian is hearing impaired B = Dalmatian is blue eyed Probabilities of hearing impairment and blue eyes among Dalmatian dogs. HI = Dalmatian is hearing impaired B = Dalmatian is blue eyed Neither HI nor B 0.66 B and not HI 0.06 HI and B 0.05 HI and not B 0.23 HI and B are not disjoint, because some dogs are both. P(HI)=0.23+0.05=0.28 P(B)=0.06+0.05=0.211 P(HI or B)=0.28+0.11-0.05=0.28+0.05+0.06=1-0.66=0.44 (all three methods are equally valid) Are the traits HI and B disjoint? Find the following: P (HI) = P (B) = P (HI or B) =

Continuous random variables Continuous sample spaces contain an infinite number of events. We use density curves to model continuous probability distributions. They assign probabilities over the range of values making up the sample space.

Continuous probabilities are assigned for intervals Events are defined over intervals of values. The total area under a density curve represents the whole population (sample space) and equals 1 (100%). Probabilities are computed as areas under the corresponding portion of the density curve for the chosen interval. The probability of an event being equal to a single numerical value is zero when the sample space is continuous. The density curve represents both the population distribution and the probability model. That is, the probability of an event corresponds to the frequency of that event in the population. P(single value) = 0 because the corresponding area under the density curve is zero.

P(y ≤ 0.5 or y > 0.8) = P(y ≤ 0.5) + P(y > 0.8) = 0.7 Let Y be a continuous random variable with a uniform distribution as illustrated here. We compute various probabilities as the corresponding areas under this uniform distribution. Height = 1 y P(0 ≤ y ≤ 0.5) = 0.5 P(0 < y < 0.5) = 0.5 P(0 ≤ y < 0.5) = 0.5 P(y = 0.5) = 0 P(y ≤ 0.5) = 0.5 P(y > 0.8) = 0.2 P(y ≤ 0.5 or y > 0.8) = P(y ≤ 0.5) + P(y > 0.8) = 0.7

Risk and odds In the health sciences, probability concepts are often expressed in terms of risk and odds. The risk of an undesirable outcome of a random phenomenon is the probability of that undesirable outcome. risk(event A) = P(event A) The odds of any outcome of a random phenomenon is the ratio of the probability of that outcome over the probability of that outcome not occurring. odds(event A) = P(event A) / [1 − P(event A)]

risk of sickle-cell anemia = 0.25. Sickle-cell anemia is a serious, inherited blood disease affecting the shape of red blood cells. Individuals carrying only one copy of the defective gene (“sickle-cell trait”) are generally healthy but may pass on the gene to their offspring. If a couple learns from prenatal tests that they both carry the sickle-cell trait, genetic laws of inheritance tell us that there is a 25% chance that they could conceive a child who will suffer from sickle-cell anemia. What are the corresponding risk and odds? The risk of conceiving a child who will suffer from sickle-cell anemia is the probability, so risk of sickle-cell anemia = 0.25. The odds is the ratio of two probabilities, so odds of sickle-cell anemia = 0.25/(1 − 0.25) = 0.333, which can also be written as odds of 1 to 3 (1:3).