5.3B Conditional Probability and Independence Multiplication Rule for Independent Events AP Statistics

Conditional Probability and Independence When knowledge that one event has happened does not change the likelihood that another event will happen, we say that the two events are independent. Two events A and B are independent if the occurrence of one event does not change the probability that the other event will happen. In other words, events A and B are independent if P(A | B) = P(A) and P(B | A) = P(B). When events A and B are independent, we can simplify the general multiplication rule since P(B| A) = P(B). Multiplication rule for independent events If A and B are independent events, then the probability that A and B both occur is P(A ∩ B) = P(A) • P(B)

Mutually Exclusive vs. Independent Select one card from a standard deck and define the events A: the card is red and B: the card is a club. Because there are no red clubs, the events are mutually exclusive. However, because P(A) = 0.5 and P(A|B) = 0, these events are not independent. Select one card from a standard deck and define the events A: the card is red and B: the card is a 7. Because there are two red 7s, A and B are not mutually exclusive. However, because P(A) = 0.5 and P(A|B) = 0.5, the events are independent.

Multiplication Rule for Independent Events Following the Space Shuttle Challenger disaster, it was determined that the failure of O-ring joints in the shuttle’s booster rockets was to blame. Under cold conditions, it was estimated that the probability that an individual O-ring joint would function properly was 0.977. Assuming O-ring joints succeed or fail independently, what is the probability all six would function properly? P( joint 1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK) By the multiplication rule for independent events, this probability is: P(joint 1 OK) · P(joint 2 OK) · P (joint 3 OK) • … · P (joint 6 OK) = (0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87 There’s an 87% chance that the shuttle would launch safely under similar conditions (and a 13% chance that it wouldn’t).

The Probability of “At Least One” Many people who come to clinics to be tested for HIV, the virus that causes AIDS, don’t come back to learn the test results. Clinics now use “rapid HIV tests” that give results while the client waits. In a clinic in Malawi, for example, the use of rapid tests increased the percent of clients who learned their tests results from 69% to 99.7%. The trade-off for fast results is that the rapid tests are less accurate than the slower laboratory tests. Applied to people who have no HIV antibodies, one rapid test has a probability of about 0.004 of predicting a false positive (that is, falsely indicating that antibodies are present when they are not). If a clinic tests 200 randomly selected people who are free of HIV antibodies, what are the chances that at least one false positive will occur?

The Probability of “At Least One” Are the test results independent from one patient to the next? It is reasonable to assume so. Since we have 200 independent events, each with a probability 0.004, “at least one” combines many different outcomes… exactly one positive exactly two positives exactly three positives….etc That’s a lot of different possibilities to calculate and add up (and could you imagine the tree diagram it would take?). Instead, let’s use our knowledge that all probabilities must add up to….1. So,

The Probability of “At Least One” Therefore….

The Probability of “At Least One” There is more than a 50% chance that at least 1 of the 200 people will test positive for HIV, even though no one has the virus.

Homework 5.3B pg. 335-337, #81, 83, 85, 89, 91, 93, 95, 97-99