The Multiplication Rules

4 CHAPTER The Multiplication Rules Introduction The previous chapter showed how the addition rules could be used to solve problems in probability. This chapter will show you how to use the multiplication rules to solve many problems in probability. In addition, the concept of independent and dependent events will be introduced. 56 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. Independent and Dependent Events The multiplication rules can be used to find the probability of two or more events that occur in sequence. For example, we can find the probability of selecting three jacks from a deck of cards on three sequential draws. Before explaining the rules, it is necessary to differentiate between independent and dependent events. Two events, A and B, are said to be independent if the fact that event A occurs does not affect the probability that event B occurs. For example, if a coin is tossed and then a die is rolled, the outcome of the coin in no way affects or changes the probability of the outcome of the die. Another example would be selecting a card from a deck, replacing it, and then selecting a second card from a deck. The outcome of the first card, as long as it is replaced, has no effect on the probability of the outcome of the second card. On the other hand, when the occurrence of the first event in some way changes the probability of the occurrence of the second event, the two events are said to be dependent. For example, suppose a card is selected from a deck and not replaced, and a second card is selected. In this case, the probability of selecting any specific card on the first draw is 1 52 , but since this card is not replaced, the probability of selecting any other specific card on the second draw is 1 51 , since there are only 51 cards left. Another example would be parking in a no parking zone and getting a parking ticket. Again, if you are legally parked, the chances of getting a parking ticket are pretty close to zero (as long as the meter does not run out). However, if you are illegally parked, your chances of getting a parking ticket dramatically increase. PRACTICE Determine whether the two events are independent or dependent. 1. Tossing a coin and selecting a card from a deck 2. Driving on ice and having an accident 3. Drawing a ball from an urn, not replacing it, and then drawing a second ball 4. Having a high I.Q. and having a large hat size 5. Tossing one coin and then tossing a second coin CHAPTER 4 The Multiplication Rules 57 ANSWERS 1. Independent. Tossing a coin has no effect on drawing a card. 2. Dependent. In most cases, driving on ice will increase the probability of having an accident. 3. Dependent. Since the first ball is not replaced before the second ball is selected, it will change the probability of a specific second ball being selected. 4. Independent. To the best of the author’s knowledge, no studies have been done showing any relationship between hat size and I.Q. 5. Independent. The outcome of the first coin does not influence the outcome of the second coin. Multiplication Rule I Before explaining the first multiplication rule, consider the example of tossing two coins. The sample space is HH, HT, TH, TT. From classical probability theory, it can be determined that the probability of getting two heads is 1 4 , since there is only one way to get two heads and there are four outcomes in the sample space. However, there is another way to determine the probability of getting two heads. In this case, the probability of getting a head on the first toss is 1 2 , and the probability of getting a head on the second toss is also 1 2 . So the probability of getting two heads can be determined by multiplying 1 2 Á 1 2 ¼ 1 4 This example illustrates the first multiplication rule. Multiplication Rule I: For two independent events A and B, PðA and BÞ¼ PðAÞÁPðBÞ. In other words, when two independent events occur in sequence, the prob- ability that both events will occur can be found by multiplying the probabil- ities of each individual event. The word and is the key word and means that both events occur in sequence and to multiply. EXAMPLE: A coin is tossed and a die is rolled. Find the probability of getting a tail on the coin and a 5 on the die. SOLUTION: Since PðtailÞ¼ 1 2 and Pð5Þ¼ 1 6 ; Pðtail and 5Þ¼PðtailÞÁPð5Þ¼ 1 2 Á 1 6 ¼ 1 12 . Note that the events are independent. CHAPTER 4 The Multiplication Rules 58 The previous example can also be solved using classical probability. Recall that the sample space for tossing a coin and rolling a die is H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6 Notice that there are 12 outcomes in the sample space and only one outcome is a tail and a 5; hence, P(tail and 5) ¼ 1 12 . EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls. A ball is selected at random and its color is noted. Then it is replaced and another ball is selected and its color is noted. Find the probability of each of these: a. Selecting 2 blue balls b. Selecting a blue ball and then a red ball c. Selecting a green ball and then a blue ball SOLUTION: Since the first ball is being replaced before the second ball is selected, the events are independent. a. There are 5 blue balls and a total of 10 balls; therefore, the probability of selecting two blue balls with replacement is P(blue and blue) ¼ PðblueÞÁPðblueÞ ¼ 5 10 Á 5 10 ¼ 25 100 ¼ 1 4 b. There are 5 blue balls and 2 red balls, so the probability of selecting a blue ball and then a red ball with replacement is Pðblue and redÞ¼PðblueÞÁPðredÞ ¼ 5 10 Á 2 10 ¼ 10 100 ¼ 1 10 CHAPTER 4 The Multiplication Rules 59 c. There are 3 green balls and 5 blue balls, so the probability of selecting a green ball and then a blue ball with replacement is Pðgreen and blueÞ¼PðgreenÞÁPðblueÞ ¼ 3 10 Á 5 10 ¼ 15 100 ¼ 3 20 The multiplication rule can be extended to 3 or more events that occur in sequence, as shown in the next example. EXAMPLE: A die is tossed 3 times. Find the probability of getting three 6s. SOLUTION: When a die is tossed, the probability of getting a six is 1 6 ; hence, the probabil- ity of getting three 6s is Pð6 and 6 and 6Þ¼Pð6ÞÁPð6ÞÁPð6Þ ¼ 1 6 Á 1 6 Á 1 6 ¼ 1 216 Another situation occurs in probability when subjects are selected from a large population. Even though the subjects are not replaced, the probability changes only slightly, so the change can be ignored. Consider the next example. EXAMPLE: It is known that 66% of the students at a large college favor building a new fitness center. If two students are selected at random, find the probability that all of them favor the building of a new fitness center. SOLUTION: Since the student population at the college is large, selecting a student does not change the 66% probability that the next student selected will favor the building of a new fitness center; hence, the probability of selecting two students who both favor the building of a new fitness center is (0.66)(0.66) ¼ 0.4356 or 43.56%. CHAPTER 4 The Multiplication Rules 60 PRACTICE 1. A card is drawn from a deck, then replaced, and a second card is drawn. Find the probability that two kings are selected. 2. If 12% of adults are left-handed, find the probability that if three adults are selected at random, all three will be left-handed. 3. If two people are selected at random, find the probability that both were born in August. 4. A coin is tossed 4 times. Find the probability of getting 4 heads. 5. A die is rolled and a card is selected at random from a deck of 52 cards. Find the probability of getting an odd number on the die and a club on the card. ANSWERS 1. The probability that 2 kings are selected is Pðking and kingÞ¼PðkingÞÁPðkingÞ¼ 4 1 52 13 Á 4 1 52 13 ¼ 1 169 2. The probability of selecting 3 adults who are left-handed is (0.12)(0.12)(0.12) ¼ 0.001728. 3. Each person has approximately 1 chance in 12 of being born in August; hence, the probability that both are born in August is 1 12 Á 1 12 ¼ 1 144 : 4. The probability of getting 4 heads is 1 2 Á 1 2 Á 1 2 Á 1 2 ¼ 1 16 . 5. The probability of getting an odd number on the die is 3 6 ¼ 1 2 , and the probability of getting a club is 13 52 ¼ 1 4 ; hence, the P(odd and club) ¼ PðoddÞÁPðclubÞ¼ 1 2 Á 1 4 ¼ 1 8 . Multiplication Rule II When two sequential events are dependent, a slight variation of the multiplication rule is used to find the probability of both events occurring. For example, when a card is selected from an ordinary deck of 52 cards the CHAPTER 4 The Multiplication Rules 61 probability of getting a specific card is 1 52 , but the probability of getting a specific card on the second draw is 1 51 since 51 cards remain. EXAMPLE: Two cards are selected from a deck and the first card is not replaced. Find the probability of getting two kings. SOLUTION: The probability of getting a king on the first draw is 4 52 and the probability of getting a king on the second draw is 3 51 , since there are 3 kings left and 51 cards left. Hence the probability of getting 2 kings when the first card is not replaced is 4 52 Á 3 51 ¼ 12 2652 ¼ 1 221 . When the two events A and B are dependent, the probability that the second event B occurs after the first event A has already occurred is written as P(B | A). This does not mean that B is divided by A; rather, it means and is read as ‘‘the probability that event B occurs given that event A has already occurred.’’ P(B | A) also means the conditional probability that event B occurs given event A has occurred. The second multiplication rule follows. Multiplication Rule II: When two events are dependent, the probability of both events occurring is PðA and BÞ¼PðAÞÁPðB j AÞ EXAMPLE: A box contains 24 toasters, 3 of which are defective. If two toasters are selected and tested, find the probability that both are defective. SOLUTION: Since there are 3 defective toasters out of 24, the probability that the first toaster is defective is 3 24 ¼ 1 8 . Since the second toaster is selected from the remaining 23 and there are two defective toasters left, the probability that it is defective is 2 23 . Hence, the probability that both toasters are defective is PðD 1 and D 2 Þ¼PðD 1 ÞÁPðD 2 jD 1 Þ¼ 3 1 24 8 4 Á 2 1 23 ¼ 1 92 CHAPTER 4 The Multiplication Rules 62 EXAMPLE: Two cards are drawn without replacement from a deck of 52 cards. Find the probability that both are queens. SOLUTION: PðQ and QÞ¼PðQÞÁPðQjQÞ ¼ 4 52 Á 3 51 ¼ 1 221 This multiplication rule can be extended to include three or more events, as shown in the next example. EXAMPLE: A box contains 3 orange balls, 3 yellow balls, and 2 white balls. Three balls are selected without replacement. Find the probability of selecting 2 yellow balls and a white ball. SOLUTION: Pðyellow and yellow and whiteÞ¼ 3 8 Á 2 7 Á 2 6 ¼ 3 1 8 4 Á 2 1 7 Á 2 1 6 1 ¼ 1 28 Remember that the key word for the multiplication rule is and. It means to multiply. When two events are dependent, the probability that the second event occurs must be adjusted for the occurrence of the first event. For the mathematical purist, only one multiplication rule is necessary for two events, and that is PðA and BÞ¼PðAÞÁPðB j AÞ: The reason is that when the events are independent PðBjAÞ¼PðBÞ since the occurrence of the first event A has no effect on the occurrence of the second event B. CHAPTER 4 The Multiplication Rules 63 PRACTICE 1. In a study, there are 8 guinea pigs; 5 are black and 3 are white. If 2 pigs are selected without replacement, find the probability that both are white. 2. In a classroom there are 8 freshmen and 6 sophomores. If three students are selected at random for a class project, find the probabil- ity that all 3 are freshmen. 3. Three cards are drawn from a deck of 52 cards without replacement. Find the probability of getting 3 diamonds. 4. A box contains 12 calculators of which 5 are defective. If two calcu- lators are selected without replacement, find the probability that both are good. 5. A large flashlight has 6 batteries. Three are dead. If two batteries are selected at random and tested, find the probability that both are dead. ANSWERS 1. P(white and white) ¼ 3 8 4 Á 2 1 7 ¼ 3 28 2. P(3 freshmen) ¼ 8 14 2 Á 7 1 13 Á 6 1 12 2 ¼ 2 13 3. P(3 diamonds) ¼ 13 1 52 4 Á 12 51 Á 11 50 ¼ 11 850 4. P(2 good) ¼ 7 12 2 Á 6 1 11 ¼ 7 22 5. P(2 batteries dead) ¼ 3 6 3 Á 2 1 5 ¼ 3 15 ¼ 1 5 Conditional Probability Previously, conditional probability was used to find the probability of sequential events occurring when they were dependent. Recall that P(B|A) means the probability of event B occurring given that event A has already occurred. Another situation where conditional probability can be used is when additional information about an event is known. Sometimes it might be CHAPTER 4 The Multiplication Rules 64 known that some outcomes in the sample space have occurred or that some outcomes cannot occur. When conditions are imposed or known on events, there is a possibility that the probability of the certain event occurring may change. For example, suppose you want to determine the probability that a house will be destroyed by a hurricane. If you used all houses in the United States as the sample space, the probability would be very small. However, if you used only the houses in the states that border the Atlantic Ocean as the sample space, the probability would be much higher. Consider the following examples. EXAMPLE: A die is rolled; find the probability of getting a 4 if it is known that an even number occurred when the die was rolled. SOLUTION: If it is known that an even number has occurred, the sample space is reduced to 2, 4, or 6. Hence the probability of getting a 4 is 1 3 since there is one chance in three of getting a 4 if it is known that the result was an even number. EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if it is known that the sum of the spots on the dice was less than six. SOLUTION: There are 2 ways to get a sum of 3. They are (1, 2) and (2, 1), and there are 10 ways to get a sum less than six. They are (1, 1), (1, 2), (2, 1), (3, 1), (2, 2), (1, 3), (1, 4), (2, 3), (3, 2), and (4, 1); hence, P(sum of 3|sum less than 6) ¼ 2 10 ¼ 1 5 . The two previous examples of conditional probability were solved using classical probability and reduced sample spaces; however, they can be solved by using the following formula for conditional probability. The conditional probability of two events A and B is PðAjBÞ¼ PðA and BÞ PðBÞ : P(A and B) means the probability of the outcomes that events A and B have in common. The two previous examples will now be solved using the formula for conditional probability. EXAMPLE: A die is rolled; find the probability of getting a 4, if it is known that an even number occurred when the die was rolled. CHAPTER 4 The Multiplication Rules 65 [...]... using one of the multiplication rules When two events are independent, the probability that the first event occurs does not affect or change the probability of the second event occurring If the events are independent, multiplication rule I is used When the two events are dependent, the probability of the second event occurring is changed after the first event occurs If the events are dependent, multiplication. .. 36 5 2 5 The problem can also be solved by looking at the reduced sample space There are two possible ways one die is a 6 and the sum of the dice is 8 There are five ways to get a sum of 8 Hence, the probability is 2 5 ¼ 2 There are 3 ways to get one tail: HT, TH, and TT There is one way to get two tails; hence, the probability of getting two tails given that one of the coins is a tail is 1 The problem... family visits the Sand Crab Water Park if it is known that they have already visited Rainbow Gardens Amusement Park 5 Three dice are rolled Find the probability of getting three twos if it is known that the sum of the spots of the three dice was six 67 CHAPTER 4 The Multiplication Rules 68 ANSWERS 1 There are five ways to get a sum of 8 They are (6, 2), (2, 6), (3, 5), (5, 3), and (4, 4) There are two... on a tail When asked why they would select a tail, they would probably respond that a tail was ‘‘due’’ according to the ‘‘law of averages.’’ In reality, however, the probability of getting a head on the CHAPTER 4 The Multiplication Rules tenth toss is 1, and the probability of getting a tail on the tenth toss is 1, so 2 2 it doesn’t really matter since the probabilities are the same A coin is an inanimate... doesn’t know that in the long run, the number of heads and the number of tails should balance out So does that make the law of averages wrong? No You see, there’s a big difference between asking the question, ‘‘What is the probability of getting 10 heads if I toss a coin ten times?’’ and ‘‘If I get 9 heads in a row, what is the probability of getting a head on the tenth toss?’’ The answer to the first 1 question... replacement, what is the probability that both will be blue? 16 81 1 b 6 1 c 2 1 d 15 a CHAPTER 4 The Multiplication Rules 10 If two people are selected at random from the phone book of a large city, find the probability that they were both born on a Sunday 1 7 1 b 49 2 c 365 2 d 7 a 11 The numbers 1 to 12 are placed in a hat, and a number is selected What is the probability that the number is 4 given... ¼ 5 ¼ CHAPTER 4 The Multiplication Rules EXAMPLE: When two dice were rolled, it is known that the sum was an even number In this case, find the probability that the sum was 8 SOLUTION: When rolling two dice, there are 18 outcomes in which the sum is an even number There are 5 ways to get a sum of 8; hence, P(sum of 8|sum is 8 even) is 18 or 4 9 EXAMPLE: In a large housing plan, 35% of the homes have...CHAPTER 4 The Multiplication Rules 66 SOLUTION: P(A and B) is the probability of getting a 4 and an even number at the same time Notice that there is only one way to get a 4 and an even number the outcome 4 Hence P(A and B) ¼ 1 Also P(B) is the probability of getting an 6 even number which is 3 ¼ 1 Now 6 2 PðA and BÞ PðBÞ 1 ¼ 6 1 2 1 1 ¼ Ä 6 2 PðAjBÞ ¼ 1 21 Á 63 1 1 ¼ 3 Notice that the answer is the same... tossed Find the probability of getting two tails if it is known that one of the coins is a tail 3 A card is selected from a deck Find the probability that it is an ace given that it is a black card 4 The probability that a family visits the Sand Crab Water Park and the Rainbow Gardens Amusement Park is 0.20 The probability that a family visits the Rainbow Gardens Amusement Park is 0.80 Find the probability... 32, so it is more likely that the next child will be a girl However, after each child is born, the probability that the next child is a girl (or a boy for that matter) is about 1 The law of 2 averages is not appropriate here My wife’s aunt had seven girls before the first boy was born Also, in the Life Science Library’s book entitled Mathematics, there is a photograph of the Landon family of Harrison, . second card. On the other hand, when the occurrence of the first event in some way changes the probability of the occurrence of the second event, the two events. influence the outcome of the second coin. Multiplication Rule I Before explaining the first multiplication rule, consider the example of tossing two coins. The