4.4-Multiplication Rule: Basics

Size: px

Start display at page:

Download "4.4-Multiplication Rule: Basics"

Bartholomew Dennis
6 years ago
Views:

1 .-Multiplication Rule: Basics The basic multiplication rule is used for finding P (A and, that is, the probability that event A occurs in a first trial and event B occurs in a second trial. If the outcome of the first event A somehow affects the probability of the second event B, it is important to adjust the probability of B to reflect the occurrence of event A. Informal Multiplication Rule: Let P (A and P (event A occurs in a first trial and event B occurs in a second trial). Then, the informal multiplication rule is stated as: P ( A and In the multiplication rule, the word and in P (A and suggests multiplication. Multiply P ( and P (, but be sure that the probability of event B takes into account the previous occurrence of event A. The derivation of this rule can be easily seen be looking at the following example. Example: A quiz has two questions. The first is a true/false question and the second is multiple choice question with five possible answers (a, b, c, d, and e). What is the probability of guessing at both questions and getting them both correct? Solution: To determine the answer we need to first determine the sample space. To do this we can make a tree diagram. A tree diagram is a picture of the possible outcomes of a procedure, shown as line segments emanating from one starting point. These diagrams are sometimes helpful in determining the number of possible outcomes in a sample space, if the number of possibilities is not too large. This figure summarizes the possible outcomes for a true/false question followed by a multiple choice question. Note that there are 0 possible outcomes in the sample space. Because only one of these ten possible combinations will give the correct answers, the probability of a favorable outcome is P (guessing correctly) 0. To put this result in more mathematical terms, let event A be guessing correctly on the first question and event B guessing correctly on the second. From the above diagram P ( and P (. 2 Therefore, 2 0

2 Example: Find the probability of correctly answering the first questions on a multiple choice test if random guesses are made and each question has 6 possible answers. Solution: The probability of answering each question is 6. If we let event A answering question correctly, event B answering question 2 correctly, etc. then we have the following: P ( A and B and C and D and E ) C) D) E) Example: When a pair of dice are rolled there are 36 different possible outcomes: -, -2, If a pair of dice are rolled times, what is the probability of getting a sum of every time? Round to eight decimal places. Solution: The favorable outcomes are; and, 2 and 3, and, and 3 and 2, so there are only four possible favorable outcomes out of 36 in the sample space. Therefore the probability on any single roll of the two dice will be P(sum of ) 9. If we let event A be rolling a sum of on the first roll, etc. then using the multiplication rule, we have: P ( A and B and C and D ) C) D) Example: A study conducted at a certain college shows that % of the school's graduates find a job in their chosen field within a year after graduation. Find the probability that randomly selected graduates all find jobs in their chosen field within a year of graduating. Round to the nearest thousandth if necessary. Solution: The probability of a favorable outcome is P ( finding a job) 0.. Let event A be the first random student finds a job, event B is the second random student finds a job, etc. then using the multiplication rule we have. P ( A and B and C and D and E ) C) D) E) (0.)(0.)(0.)(0.)(0.) (0.) 0.03

3 Conditional Probability: In some situations, the second event will be affected by the outcome of the first event. In this case, the probability of the second event must take into account the fact that the first event has already occurred. This is easily seen in the following example: Example: You are dealt two cards successively (without replacement) from a shuffled deck of 2 playing cards. Find the probability that the first card is a King and the second card is a queen. Express your answer as a simplified fraction. Solution: let event A drawing a King and event B drawing a Queen. When determining the sample space for event B, we must take into account that event A has already occurred. We then have the following probabilities: P ( 2 3 P (. The sample space for event B takes into account the fact that one card had already been removed from the deck, leaving only cards. We now use the multiplication rule: 3 Notation for Conditional Probability: P (B represents the probability of event B occurring after it is assumed that event A has already occurred (read B A as B given A. ). We can modify the previous multiplication rule as follows. B Example: You are dealt two cards successively (without replacement) from a shuffled deck of 2 playing cards. Find the probability that both cards are black. Express your answer as a simplified fraction. Solution: Let event A picking the first black card and event B picking the second black card. Because the cards are drawn without replacement the outcome of B is dependent on the outcome of A and is therefore a conditional probability problem. Use the formula: B 663 The probabilities are: 26 P ( 2 2 P ( B 2 2 B 2 Note: If this problem was done with replacement the probability would be:

Independent and Dependent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other.

4 Independent and Dependent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are similarly independent if the occurrence of any does not affect the probabilities of the occurrence of the others.) If A and B are not independent, they are said to be dependent. Two events are dependent if the occurrence of one of them affects the probability of the occurrence of the other, but this does not necessarily mean that one of the events is a cause of the other. Formal Multiplication Rule: B If A and B are independent events, then Caution: When applying the multiplication rule, always consider whether the events are independent or dependent, and adjust the calculations accordingly. For each of the following examples, identify the events as dependent or independent. Then, determine the probability. Example: In a homicide case 7 different witnesses picked the same man from a line up. The line up contained men. If the identifications were made by random guesses, find the probability that all 7 witnesses would pick the same person. Solution: These events are independent. The probability of each event is, therefore using the multiplication rule we have: P (all picking same person) 7 782

5 Example: A sample of different calculators is randomly selected from a group containing 6 that are defective and 26 that have no defects. What is the probability that all four of the calculators selected are defective? Round to four decimal places. Solution: These are dependent events. The probability of each successive event is dependent on what occurred in the previous event. 6 P( 72 P( 7 P(C) 70 3 P(D) 69 P(all calculators are defective) 0.86 Example: A bin contains 78 light bulbs of which 9 are defective. If 3 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. Round to the nearest thousandth if necessary. Solution: These events are independent because of the replacement. Consequently, each event will have a 69 probability of and the probability of selecting three good ones is Treating Dependent Events as Independent: The % Guideline Some calculations are cumbersome, but they can be made manageable by using the common practice of treating events as independent when small samples are drawn from large populations. In such cases, it is rare to select the same item twice. If a sample size is no more than % of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent). Example: A batch of 00,000 heart pacemakers includes 99,90 that are good and 0 that are bad. Find the probability that 20 randomly selected pacemakers will be good. Solution: There is no replacement so these events are dependent. Because the sample size of 20 is less than % of the total population, we can treat these events as independent rather than dependent. The 99,90 probability of selecting a good pacemaker is. 00,000 Therefore, 99,90 P(all 20 pacemakers are good) ,000 Note: If this problem was treated as dependent events, the calculation would have been much more complex. P (all 20 pacemakers are good) 99,90 00, ,99 99,98 Λ 99,999 99,998 99,93 99,98

Lecture Slides. Elementary Statistics Eleventh Edition. by Mario F. Triola. and the Triola Statistics Series 4.1-1

Lecture Slides. Elementary Statistics Eleventh Edition. by Mario F. Triola. and the Triola Statistics Series 4.1-1 Lecture Slides Elementary Statistics Eleventh Edition and the Triola Statistics Series by Mario F. Triola 4.1-1 4-1 Review and Preview Chapter 4 Probability 4-2 Basic Concepts of Probability 4-3 Addition