Independent EventsTwo events are said to be independent if the outcome of oneof them does not influence the other. For example, in sportingevents, the outcomes of different games are usually consideredindependent even though that may not be true in a completelystrict and literal sense.The multiplication rule for independent events says that if A andB are independent,P (A and B) = P (A) × P (B).1Example: A football pundit states that the probability that UNCwill beat NC State is 0.4, while the probability that UNC willbeat Duke is 0.8. What is the probability that1. UNC wins both games?2. UNC wins at least one game?3. UNC loses both games?2Solution:1. If A is the event “UNC beats State” and B is the event “UNCbeats Duke”, and if we assume these are independent events,then the probability of A and B is 0.4 × 0.8 = 0.32.2. Apply the Law of Addition:P (A or B) = P (A)+P (B)−P (A and B) = 0.4+0.8−0.32 = 0.88.3. Apply the Law of Complementary Events: “UNC loses bothgames” is the complement of “UNC wins at least one game”,so its probability is 1 − 0.88 = 0.12.3Warning:Don’t confuse the notions of “independent events” and “disjointevents”. Independence means that the outcome of one eventdoes not influence the outcome of the other. Disjoint meansthat if one event occurs then the other cannot occur — the veryopposite of independence!4Conditional ProbabilitiesConsider the example (page 239 of text, referring to the Wim-bledon tennis tournament),A: “Federer misses his first serve”B: “Federer misses his second serve”We are told that Federer misses his first serve 44% of the time,and that of all the times he misses his first serve, he also misseshis second serve 2% of the time.What, then, is the probability he has a double fault?Logically, the answer is 2% of 44%, or 0.02×0.44, which is about0.009.5Now let us rephrase this in the language of conditional probability.We are told that the event A occurs 44% of the time, or in otherwords P (A) = 0.44.We are also told that, given that A has occurred, the event Boccurs 2% of the time. This is written in probability notation asP (B | A) = 0.02.The left hand side is read as the probability of B given A. In thisparticular context, it would not make sense to talk about theprobability of B given Ac, though in other contexts, that wouldmake sense (e.g. free throws in basketball).6The law of multiplication for conditional probabilities saysP (A and B) = P (A) × P (B | A).Note that if we just interchange the role of A and B, we also getP (A and B) = P (B) × P (A | B).Finally, if A and B are independent, we get P (A | B) = P (A)and P (B | A) = P (B) — that formalizes what is meant by sayingthat the outcome of one event does not influence the outcomeof the other. But in that case, either of the last two formulasreduces toP (A and B) = P (A) × P (B)as in our earlier formulation of the multiplication rule for inde-pendent events.7Here is another (more complicated) example.Consider the game in which a player tosses a die twice, and wewant to calculate the probability that the total of the two tossesis at least 10. Define the eventsA: The first throw is a 6.B: The first throw is a 5.C: The first throw is a 4.D: The total of the two throws is at least 10.Note that if the first throw is less than 4, it’s impossible for thetotal to be 10 or higher. SoP (D) = P (A and D) + P (B and D) + P (C and D). (1)8Now P (A) =16. Given that A has occurred, D will occur if thesecond throw produces any of 4, 5 or 6, and the probability ofone of those outcomes is36or12. So we haveP (A) =16, P (D | A) =12, P (A and D) =16×12=112.SimilarlyP (B) =16, P (D | B) =13, P (B and D) =16×13=118,P (C) =16, P (D | C) =16, P (C and D) =16×16=136.Therefore, (1) leads us toP (D) =112+118+136=16giving the same answer as in our earlier calculation.9Tree DiagramsA diagnostic test for a certain type of cancer has a 98% chanceof giving the correct outcome. Among all patients who take thisparticular test, 0.3% have the cancer in question. You take thistest and the result comes back positive. What is the probabilitythat you actually have cancer?(A) 98%(B) 86%(C) 41%(D) 13%10Solution: Draw a tree diagramNO CANCER0.997CANCER0.003POSITIVE TEST0.02NEGATIVE TEST0.98POSITIVE TEST0.98NEGATIVE TEST0.020.019940.977060.002940.0000611If A is the event “patient has cancer” and B is the event “patienthas positive test”, thenP (A and B) = 0.00294,P (B) = 0.00294 + 0.01994 = 0.02288,P (A | B) =0.002940.02288= 0.1285.So the answer is about 13%. To many people encountering thisissue for the first time, it is very surprising that the chance ofactually having the disease, given a positive test, is so low. Butit illustrates the problem created by “false positives” in manydiagnostic