Transcript Slide 1
General rules of probability BPS chapter 12 © 2006 W.H. Freeman and Company 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 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. Two events are disjoint if they have no outcomes in common Events A and B are disjoint. and can never happen together. Events A and B are NOT disjoint. If you flip a coin once, it may turn out head or tail. However, you cannot obtain both head and tail on the same flip. Head and tail are disjoint events. If a couple gets a child, the child could be a boy or a girl. The child cannot be both boy and girl. Boy and girl are disjoint events. 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 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 Conditional probability Conditional probabilities reflect 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) P( A and B) P( B | A) P( A) If A and B are independent, P(B | A) = P(B). 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. Probability trees Conditional probabilities may 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. To find out how often adults use Internet chat rooms, a survey of adult Internet users was conducted and this tree diagram was constructed. The tree diagram shows the chat room usage of three adult age groups. Internet user 0.47 P(chatting) = 0.136 + 0.099 + 0.017 = 0.252 About 25% of all adult Internet users visit chat rooms. Breast cancer screening It is recommended that women in their 20s get screened for breast cancer. Not all positive results in the screening accurately detect cancer. If woman receives a positive test result, what is the probability that she has breast cancer? Diagnosis sensitivity 0.8 Disease incidence 0.0004 Positive Cancer 0.2 Mammography 0.9996 0.1 Negative False negative False positive Positive No cancer Incidence of breast cancer among women ages 20–30 0.9 Diagnosis specificity Negative Mammography performance She could either have a positive test and have breast cancer, or have a positive test but not have cancer (false positive). Diagnosis sensitivity 0.8 Disease incidence Positive Cancer 0.0004 0.2 Mammography 0.1 0.9996 Negative False negative Positive False positive No cancer Incidence of breast cancer among women ages 20–30 0.9 Diagnosis specificity Negative Mammography performance Possible outcomes given the positive diagnosis: positive test and breast cancer, or positive test but not cancer (false positive). P(cancer | pos ) P(cancer and pos ) P(cancer and pos ) P (nocancer and pos ) 0.0004*0.8 0.3% 0.0004*0.8 0.9996*0.1 This value is called the positive predictive value, or PV+. It is an important piece of information but, unfortunately, is rarely communicated to patients.