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Objectives (BPS chapter 12)

Objectives (BPS chapter 12). General rules of probability Independence and the multiplication rule The general addition rule Conditional probability The general multiplication rule Independence Tree diagrams. Independent versus disjoint events.

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Objectives (BPS chapter 12)

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  1. Objectives (BPS chapter 12) General rules of probability • Independence and the multiplication rule • The general addition rule • Conditional probability • The general multiplication rule • Independence • Tree diagrams

  2. Independent versus disjoint events Two events are independent if the probability that one event occurs on any given trial of an experiment is not influenced in any way by the occurrence of the other event. Imagine coins spread out so that half were heads up, and half were tails up. Pick a coin at random. The probability that it is heads-up is 0.5. But, if you don’t put it back, the probability of picking up another heads-up coin is now less than 0.5. Without replacement, successive trials are not independent. In this example, the trials are independent only when you put the coin back (“sampling with replacement”) each time.

  3. The multiplication rule for independent events Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, P(A and B) = P(A) P(B) If A is the event that you roll a “1” on a red 6-sided die, and B is the event that you roll a “1” on a blue 4-sided die, what is the probability that if you roll both at the same time, you will roll a “1” on both dice? Since the two dice are independent, P(both “1”s) = P(“1” on red die) P(“1” on blue die) = (1/6) (1/4) = 1/24

  4. The general addition rule General addition rule for any two events A and B: The probability that A occurs, or B occurs, or both events occur is: P(A or B) = P(A) + P(B) – P(A and B) What is the probability of randomly drawing either an ace or a heart from a pack of 52 playing cards? There are 4 aces in the pack and 13 hearts. However, one card is both an ace and a heart. Thus: P(ace or heart) = P(ace) + P(heart) – P(ace and heart) = 4/52 + 13/52 - 1/52 = 16/52 ≈ 0.3

  5. Conditional probability Conditional probabilitiesreflect how the probability of an event can change if we know that some other event has occurred/is occurring. • Example: The probability that a cloudy day will result in rain is different if you live in Los Angeles than if you live in Seattle. • Our brains effortlessly calculate conditional probabilities, updating our “degree of belief” with each new piece of evidence. The conditional probability of event B given event A is:(provided that P(A) ≠ 0) If A and B are independent, P(B | A) =P(B).

  6. The general multiplication rule • The probability that any two events, A and B, both occur is: P(A and B) = P(A)P(B|A) This is the general multiplication rule. • If A and B are independent, then P(A and B) = P(A)P(B) A and B are independent when they have no influence on each other’s occurrence. Another way of showing this is P (B | A) = P (B) or P (A | B) = P (A). • What is the probability of randomly drawing either an ace or heart from a pack of 52 playing cards? There are four aces in the pack and thirteen hearts. • P(heart|ace) = 1/4 P(ace) = 4/52 • P(ace and heart) = P(ace)* P(heart|ace) = (4/52)*(1/4) = 1/52 • Notice that heart and ace are independent events. P(ace or heart) =P(ace)+P(heart)-P(ace and heart)=4/52+13/52-1/52=16/52

  7. Tree diagram for chat room habits for three adult age groups. 0.47 Internet user Probability trees Conditional probabilities can get complex and it is often a good strategy to build a probability tree that represents all possible outcomes graphically and assigns conditional probabilities to subsets of events. P(chatting) = 0.136 + 0.099 + 0.017 = 0.252 About 25% of all adult Internet users visit chat rooms.

  8. Diagnosis sensitivity 0.8 Disease incidence Positive Cancer 0.0004 False negative Negative 0.2 Mammography 0.1 False positive Positive 0.9996 No cancer Negative Incidence of breast cancer among women ages 20–30 0.9 Diagnosis specificity Mammography performance Breast cancer screening If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she has breast cancer? She could either have a positive test and have breast cancer, or have a positive test but not have cancer (false positive).

  9. Diagnosis sensitivity Disease incidence Positive Cancer False negative Negative Mammography False positive Positive No cancer Negative Diagnosis specificity 0.8 0.0004 0.2 0.1 0.9996 Incidence of breast cancer among women ages 20–30 0.9 Mammography performance Possible outcomes given the positive diagnosis: positive test and breast cancer, or positive test but not cancer (false positive). This value is called the positive predictive value, or PV+. It is an important piece of information but, unfortunately, is rarely communicated to patients.

  10. Practice Problem 1 Musical styles other than rock and pop are becoming more popular. A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both country and gospel music. • What is the probability that a randomly selected college student likes gospel but not country music?

  11. Practice Problem 1 • What is the probability that a randomly selected college student doesn’t like gospel or country music? • What is the probability that the student chosen likes gospel music given that he/she likes country music?

  12. Practice Problem 2 • You and a friend decide to have lunch at a nearby Chinese restaurant. The table below lists the six meals you and your friend ever order, along with the price of the dish, the probabilities that you will order each dish, and the probabilities your friend will order each dish. Assume that your choice of a meal has no bearing on your friend’s choice (and vice versa).

  13. Practice Problem 2 1. Let A be the event that you order a spicy lunch. Find P(A). • P(A)= P(Szechuen Chicken or Curry Shrimp) =0.1+0.3=0.4 2. Let B be the event you do not order a spicy lunch, and C be the event you order a lunch that costs less than $4.60. Find P(B and C). P (B and C) = P(Roast Pork with Bean Sprouts or Beef with Broccoli or Chicken Wings) = 0.05 + 0.1+ 0.1=0.25 3. With B and C defined in the previous part, find P(B or C). P(B or C) = P(𝐵) + P(C) − P(B and C) = 0.6 + 0.35 − 0.25 = 0.7 4. What is the probability that both you and your friend order spicy lunches? P(A and E) P(A)P(E) = 0.4 × 0.4 = 0.16

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