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Chapter 3

Chapter 3. Probability. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 2 I. SOME PROBABILITY RULES -- COMPOUND EVENTS

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Chapter 3

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  1. Chapter 3 Probability

  2. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 2 I. SOME PROBABILITY RULES -- COMPOUND EVENTS We have seen how to calculate the probability of simple events for a statistical experiment. Compound events are those which combine two or more simple events. Two connectives with which simple events are combined to produce compound events are AND and OR. It is essential to differentiate between them and to recognize special cases for each. A. The AND combination P(A and B). 1. Used when we are required to compute the probability of two events which occur together or in sequence. 2. Look for words like AND, BOTH, TOGETHER, and phrases which combine two traits, like KING OF SPADES (king AND spade), AGGRESSIVE SALESPERSON (aggressive AND salesperson), MALE NURSE (male AND nurse), overworked teacher, paranoid schizophrenic, etc. 3. To select the proper formula, we must first determine whether events A and B are dependent or independent. a) The two events are independent if P(A) does not affect P(B). 1) Roll a die and flip a coin. No matter what the die roll is, the probability that the coin will come up heads is still ½. Likewise, no matter what the coin does, the probability of rolling a 5 is still 1/6.

  3. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 2 a When the events are independent, use the multiplication rule. 1 P(A and B) = P(A) x P(B). a). P(head and 5) = P(head) x P(5) = (1/2)(1/6) = 1/12 b) The two events are dependent if P(A) affects P(B). 1) In the rule for dependent events, we must examine the probability that event B will occur under the condition that event A has already occurred. This is called conditional probability and will be denoted P(B, given A) a. Example: A jar contains 5 red marbles, 3 green, and 2 blue. Draw two marbles without replacing the first one before you draw the second. Find the probability the first marble is red and the second is green. 1. A = first marble is red 2. B = second marble is green. P(green) depends on the color of the first marble. b. When the events are dependent, use the following multiplication rule: 1. P(A and B) = P(A) x P(B given A)

  4. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 2 a) Calculate P(A) b) Assume that A has occurred. c) Make the necessary adjustments to the sample space, and calculate P(B given A). d) Multiply these two values. 1) P(red) = 5/10, or ½. a Once a red marble has been drawn, only 9 marbles left, 3 of which are green. 2) P(B given A) = 3/9 or 1/3. 3) P(A and B) = P(A) x P(B given A) = (1/2)(1/3) or 1/6. 4. For most AND situations, we will have to use intuition to determine whether two events are independent or dependent (and consequently which multiplication rule to use). However, with enough information, we can figure out whether two events are independent. a) We know that if two events are independent, then P(A and B) = P(A) x P(B). The converse of this is also true. In other words, if P(A) x P(B) = P(A and B), then the two events are independent.

  5. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 2 5. When the two events are independent, P(B) = P(B given A). a) The chance that B will occur is not influenced by whether or not A has occurred. 1) To decide whether two events are independent: a. Evaluate P(B). b. Evaluate P(B given A) c. When the two results are equal, the events are independent d. When the two results are not equal, the events are dependent.

  6. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 A. The OR combination P(A or B) 1. The question concerns the probability of either or two (or more) events occurring. When we say “A or B”, we include the possibility of just A, just B, or both A and B. 2. Look for key words like OR, EITHER, or phrases like AT LEAST ONE. 3. To select the correct formula, we must first determine whether or not the events A and B are mutually exclusive. a) A and B are mutually exclusive if they cannot both occur within the same trial or experiment. 1) If A occurs, then B cannot, and vice-versa. 2) In other words, the occurrence of A excludes the occurrence of B. 3) Example: Draw a card from a deck. a. A = card is a ten b. B = card is a king 1. Once we know that the card is a ten, it cannot be a king. The two are mutually exclusive.

  7. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 4) When the two events are mutually exclusive, use the simple addition rule: a. P(A or B) = P(A) + P(B). 1. P(A) = 4/52, or 1/13. 2. P(B) = 4/52, or 1/13. 3. P(A or B) = 1/13 + 1/13 = 2/13. b) A and B are not mutually exclusive if they can both occur within the same trial or experiment. 1) Example: Draw a card from a deck a. A = card is a ten b. B = card is a diamond. 1. The card can be both and ten and a diamond; the two events do not exclude each other. 2) When the events are not mutually exclusive, use the following addition rule: a. P(A or B) = (P(A)+P(B))-P(A and B). 1. Calculate P(A), and P(B). Add them together. 2. Calculate P(A and B).

  8. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTION 3 3. Subtract P(A and B) from the total you got in step 1. a) This is to prevent counting an event twice. 1) If the card happened to be the ten of diamonds, it would be counted as both a ten and as a diamond. Subtracting P(A and B) prevents this. b. P(A) = 4/52 There are four cards in the deck that are tens c. P(B) = 13/52 There are 13 cards in the deck that are diamonds. d. P(A and B) = 1/52. There is only one card in the deck that is both a ten and a diamond. e. P(A or B) = (4/52 + 13/52) - 1/52 = 16/52. 1. There are 16 cards in the deck that are either a ten, or a diamond, or both. One card, the ten of diamonds, is both and is counted twice. We subtract it once to keep the count accurate. f. When the two events are mutually exclusive, P(A and B) = 0. 1. The two events cannot happen together.

  9. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES Steps to solving problems 1. Determine what the events are. 2. Decide whether to use the “and” combination, the “or” combination, or the complement rule. 3. Before choosing the correct formula, ask one of the following questions: a) For the AND combination: ARE THE EVENTS INDEPENDENT? b) For the OR combination: ARE THE EVENTS MUTUALLY EXCLUSIVE? 4. After answering question #3, write down the appropriate formula. 5. Finally, answer the probability question.

  10. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES Two marbles are drawn without replacement from a bowl containing 7 red, 4 blue, and 9 yellow marbles. A. What is the probability that both are red? Events are: A = First is Red B = Second is Red This is an “AND” combination (the first is Red AND the second is Red) To determine the correct formula, we need to know “are the events independent or dependent?” The events are dependent since the first marble is not replaced before the second marble is drawn. Using the formula P(A and B) = P(A) * P(B, given A), we get: P(A) = 7/20 P(B, given A) = 6/19 (7/20)(6/19) = 42/380 = 21/190 = .1105. There is an 11.05% chance of getting two red marbles.

  11. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES Two marbles are drawn without replacement from a bowl containing 7 red, 4 blue, and 9 yellow marbles. B. What is the probability that NEITHER is red? This is an AND combination as well (The first marble is not red, AND the second marble is not red) The events are dependent, again because the first marble is not replaced before drawing the second P(First not Red) = 13/20 P(Second not Red, given First not Red) = 12/19. (13/20)(12/19) = 156/380 = 39/95 = .4105 C. What is the probability that at least one is red? The complement to “at least one” is “neither”. We can use the complement rule and get: P(at least one is red) = 1 – P(neither is red) = 1 – 39/95 = 56/95 = .5895

  12. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) A. What is the probability that it is a 4 or a 5? The events are: A = card is a 4 B = card is a 5. This is an OR combination: Either the card is a 4, OR it is a 5. The events are mutually exclusive, since a card cannot be both a 4 and a 5 at the same time. Use the simple addition rule. P(A or B) = P(A) + P(B) P(4 or 5) = 4/52 + 4/52 = 8/52 = 2/13 = .1538

  13. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) B. What is the probability that the card is neither a 4 nor a 5? This is the complement to A. The card is NOT a “4 or 5” 1 – 2/13 = 11/13 = .8461

  14. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES One card is drawn from a standard deck of 52 cards (no jokers) C. Find the probability that it is either a 4 or it is red. This is an OR combination. Are the events mutually exclusive? NO – a card can be both a 4 and red at the same time. Use the addition rule to compute the probability P(A or B) = P(A) + P(B) – P(A and B) P(A) = 4/52 and P(B) = 26/52 (There are 4 cards that are 4’s, and 26 red cards) P( A and B) = 2/52 (two red 4’s) P(4 or Red) = 4/52 + 26/52 – 2/52 = 28/52 = .5385

  15. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES Two dice are rolled and the sum of the dice is noted. A. What is the probability that the sum is a 7 or 11? This is an OR combination, and the events are mutually exclusive. Use the simple addition rule: P(A or B) = P(A) + P(B) P(7 or 11) = 6/36 + 2/36) = 8/36 = .2222

  16. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES Two dice are rolled and the sum of the dice is noted. B. What is the probability that the sum is at least ten? This is an OR combination. At least ten means 10 OR 11 OR 12. These are mutually exclusive. P(at least 10) = P(10) + P(11) + P(12) = 3/36 + 2/36 + 1/36 = 6/36 = .1667 C. What is the probability that the sum is less than 10? This is the complement to at least ten. P(less than 10) = 1 – P(at least 10) = 1 – 6/36 = 30/36 = .8333

  17. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES At Studymore University, 46% of the students are female and 13% of the females are psychology majors. 9% of all the students are psychology majors. A student is picked at random. A. Find the probability that the person is a female psychology major. This is an AND combination (a female AND a psychology major) Are the events independent? No – the probability of a female majoring in psychology is different than the probability of a student at large majoring in psychology (13% to 9%) P(female and a psychology major) = P(female) * P(psychology major, given female) P(female and a psychology major) = .46 * .13 = .0598

  18. PROBABILITY AND STATISTICS CHAPTER 3 NOTES SECTIONS 3-2 and 3-3 EXAMPLES At Studymore University, 46% of the students are female and 13% of the females are psychology majors. 9% of all the students are psychology majors. A student is picked at random. B. Find the probability that the student is a female or a psychology major This is an OR combination The events are not mutually exclusive: one can be both female and majoring in psychology at the same time. P(female or psychology major) = P(female) + P(psychology major) – P(female and psychology major) .46 + .09 - .0598 = .4902

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