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Section 4.2 Probability Rules
Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. HAWKES LEARNING SYSTEMS math courseware specialists Section 4.2 Probability Rules
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P(S) = 1, where S is the sample space of all possible outcomes
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Facts about Probability: 0 ≤ P(E) ≤ 1 P(S) = 1, where S is the sample space of all possible outcomes P(Ø) = 0, where Ø is the empty set
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For an event E and its complement Ec: P(E) = 1 – P(Ec)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules The Complement: The complement for E, denoted Ec, consists of all outcomes in the sample space that are not in E. Probability Rule for the Complement: For an event E and its complement Ec: P(E) = 1 – P(Ec) P(E) = 1 – 0.95 = 0.05
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Choose a red card out of a standard deck of cards. All 26 black cards.
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Describe the complement for each of the following events: Choose a red card out of a standard deck of cards. All 26 black cards. Out of 31 students in your statistics class, 15 are out sick with the flu. The 16 students that are not sick. In your area, 91% of phone customers use PhoneSouth. The other 9% of customers in your area who do not use PhoneSouth.
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Roll a pair of dice. What is the probability that neither die is a 3?
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Find the probability: Roll a pair of dice. What is the probability that neither die is a 3? Solution: It would be tedious to write out every combination that does not contain a 3. Using the complement, we could list the outcomes which either die contains a 3. There are 11 outcomes where at least one of the dice is a 3.
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P(E or F) = P(E) + P(F) – P(E and F)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Addition Rule: The probability that one event happens or the other event happens. Addition Rule for Probability: For two events E and F: P(E or F) = P(E) + P(F) – P(E and F) A and B A (Universal) B (Disney) Not mutually exclusive – it shares outcome
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P(E or F) = P(E) + P(F) – P(E and F)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Calculate the probability: Find the probability of choosing either a spade or a face card (king, queen, jack) out of a standard deck of cards. Solution: P(E or F) = P(E) + P(F) – P(E and F) P(spade or face card) = P(spade) + P(face card) – P( spade and face card)
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A (Universal) B (Disney) Events that share no outcome.
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Mutually Exclusive Events: Events that share no outcome. Addition Rule for Mutually Exclusive Events: If two events, E and F, are mutually exclusive then: P(E or F) = P(E) + P(F) A (Universal) B (Disney) Mutually exclusive – it shares no outcome
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P(face card or 7) = P(face card) + P(7)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Calculate the probability: What is the probability of drawing a face card or a seven from a standard deck of cards? Solution: P(E or F) = P(E) + P(F) P(face card or 7) = P(face card) + P(7)
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With repetition – outcomes may be repeated.
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Definitions: With repetition – outcomes may be repeated. Without repetition – outcomes may not be repeated. With replacement – objects are placed back into consideration for the following choice. Without replacement – objects are not placed back into consideration for the following choice. Independent events – if one event happening does not influence the probability of the other event happening. Multiplication Rule for Probability: For two independent events, E and F: P(E and F) = P(E)P(F)
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P(king and queen) = P(king) P(queen)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Calculate the probability: Choose two cards from a standard deck, with replacement. What is the probability of choosing a king and then a queen? Solution: P(E and F) = P(E)P(F) P(king and queen) = P(king) P(queen)
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HAWKES LEARNING SYSTEMS
math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Conditional Probability: When two events are not independent, the outcome of one influences the probability of the other. P(F|E) is read as “the probability of event F occurring given event E occurred first”.
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First we need to determine the number of red cards left in the deck.
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Calculate the probability: What is the probability of choosing a red card from the deck, given that the first card drawn was a diamond? Assume the cards are chosen without replacement. P(red) = 26 out of 52 Solution: First we need to determine the number of red cards left in the deck. Since a diamond is a red card and has already been chosen there are only 25 red cards left.
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P(E and F) = P(E) * P(F|E)
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Multiplication Rule for Dependent Events: For two dependent events, E and F: P(E and F) = P(E) * P(F|E) What is the probability of choosing a white shirt and then a blue shirt without replacement? Total = 10 White and 10 Blue = 20 shirts P(W and B) = P(W) * P(B|W) = = 10/20 * 10/19 = P(W and B) =
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P(face card and face card) = P(1st face card) x
HAWKES LEARNING SYSTEMS math courseware specialists Probability, Randomness, and Uncertainty 4.2 Probability Rules Calculate the probability: What is the probability of choosing two face cards in a row without replacement? Solution: P(E and F) = P(E)P(F|E) P(face card and face card) = P(1st face card) x P(2nd face card| 1st face card)
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Not Mutually Exclusive P = P(SMALL) + P(GREEN) - P(SMALL and GREEN)
20 LARGE 3 GREEN 17 WHITE 15 SMALL 12 35 TOTAL
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Probability of White, being that the first was a White marble?
W = = 12 left Total = = = 17 left P(W|W) = 12 ÷ 17 =
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P(Nickel and Dime) = P(Nickel) x P(Dime | Nickel)
P(Nickel) = 12 ÷ 65 = 64 coins left P(Dime | Nickel) = 13 ÷ 64 P(Nickel and Dime) = (12 ÷ 65) x (13 ÷ 64) =
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P(M and Fraternity)= P(M) x P(Fraternity|Male)
P(M and Fraternity)= P(M) x P(Fraternity|Male) P(M and Fraternity)= 894/1870 x 227/894 = 227/1870 =
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Copy Data from Hawkes to Excel
Paste in cell A2
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P(Not M or Not With Parents)= 1 – P(M and With Parents)
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Copy Data from Hawkes to Excel
Paste in cell A2
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