Section 8.4 – Independence 369 Section 8.4 Independence In section 8.3 we looked at many scenarios in which the known result of one event F affected the probability of another event E. These were summarized as conditional probabilities, and in each case, the probability of )|( FEP, the probability of event E given event F, was calculated using the formula(s): )()()()()|(FnFEnFPFEPFEP∩=∩= Usually, knowing F had a tangible effect on the probability of E, so that in these cases, )|()( FEPEP≠. Suppose knowing event F has no effect on the probability of E. In other words, )|()( FEPEP=. What does this mean exactly? How can this be interpreted? Consider the first example, meant to convey this idea intuitively: Example 1: A single card from a standard deck of 52 is chosen. Let E = “the card is a King” and F = “the card is a Diamond”. Calculate these probabilities: a) )(EP, the probability the card is a King. b) )(FP, the probability the card is a Diamond. c) )( FEP∩, the probability the card is a King and a Diamond. d) )|( FEP, the probability the card is a King given the card is a Diamond. e) )|( EFP, the card is a Diamond given the card is a King. Discussion: The first two can be calculated quickly. The entire sample space of 52 cards is considered. There are 4 Kings and 13 Diamonds, so for (a) the probability the card is a King is 131524)( ==EP and for (b), the probability the card is a Diamond is 415213)( ==FP For (c), the intersection FE∩ is the set of cards that are Kings and Diamonds. There is just one: the King of Diamonds, so 521)( =∩ FEPSection 8.4 – Independence 370 Now, for part (d), the probability the card is a King given it is a Diamond, we use the formula (using parts (c) and (b) within it): 131)()()|(5213521==∩=FPFEPFEP Note that this is the same as )(Ep . In other words, knowing the card is a Diamond had no effect on the probability that the card was a King. The probability remained unchanged. It’s as if the given information was not needed at all. Similarly, for part (e), the probability the card is a Diamond given it is a King, we get: 41)()()|(524521==∩=EPFEPEFP Again, this is the same as )(Fp . Again, given the fact the card was a King had no effect on the probability the card would be a Diamond. In both cases (d) and (e), we can summarize the following facts: )|()( FEPEP= and )|()( EFPFP= When these statements are true, the two events are independent of one another. Test For Independence The above statements can be used to generate a useful formula called the test for independence formula. We start with the formula for conditional probability: )()()|(FPFEPFEP∩= Assuming E and F are independent, we can replace the )|( FEP with )(EP : )()()(FPFEPEP∩= Multiplying out the denominator, we get the following formula and definition: Test for Independence Formula: Two events E and F are independent if the formula below is true: )()()( FEPFPEP∩=×. If the formula is false, then the two events are dependent (conditional).Section 8.4 – Independence 371 The formula is a True/False formula: you are testing the equality of the values. You are obliged to determine the various numerical values using other methods. The following examples illustrate this formula: Example 2: Two fair dice are rolled. Let E = “the sum is 6” and F = “both dice show the same value (doubles)”. Are the two events independent of one another? Discussion: We need to calculate the three parts of the formula individually. The easiest way is to view the entire sample space of 36 rolls and count off the elements in each event: • There are five ways to roll a six: E = { (1,5), (2,4), (3,3), (4,2), (5,1) }, so 365)(=EP . • There are six ways to roll “doubles”: F = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }, so 366)(=FP . • There is just one element that is simultaneously a sum of six and a doubles: })3,3({=∩FE , so 361)(=∩FEP . Place these three calculations into the formula and check for truth: FalseFEPFPEP ,)()()(361?366365?≠⇒=×⇒∩=× The formula is false, so the two events are not independent of one another. They are dependent. Example 3: Two fair dice are rolled. Let E = “first die shows a 1” and F = “second die shows a 2”. Are E and F independent? Discussion: As in example 2, the three individual calculations must be determined individually: • There are six was the first die can show a one: { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) }, so 366)( =EP . • There are six ways the second die can show a two: { (1,2), (2,2), (3,2), (4,2), (5,2), (6,2) }, so 366)( =FP . • There is just one way to have the first die show a 1 and the second die show a 2 at the same time: })2,1({=∩FE, so 361)( =∩ FEP . Now insert these values into the formula and check: TrueFEPFPEP ⇒=⋅⇒∩=×361?366366?)()()( The formula is true, so the two events are independent of one another. Intuitively, this should seem correct: throwing a 1 with the first die should not affect the probability of throwing a 2 with the second die. Section 8.4 – Independence 372 Do not use the Test For Independence formula as a calculation device unless you are TOLD the events are independent! It is tempting to use the formula to calculate )( FEP∩ by multiplying )(EP and )(FP when trying to show independence. This is bad circular reasoning: you cannot use what you are trying to prove. The probability )( FEP∩ must be calculated some other way. However, if the events E and F are stated to be independent as fact, then you may use the formula as a numerical calculation device. Example 4: Suppose two events A and B are independent and 45.0)(=AP and 85.0)(=BP. What is )( BAP∩? Discussion: Since the events were stated to be independent, then you may use the formula to numerically calculate )( BAp∩: 3825.0)85.0()45.0()()()()(=×=∩⇒×=∩BAPBPAPBAP Example 5: My business is serviced by UPS and FedEx. I have noticed that UPS comes to my business with a probability of 0.2 on any given day, and FedEx comes to my business with a probability of 0.35 on any given day. They come independently of one another. What’s the probability that both will show up on any given day? Discussion: Since the events are
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