Introduction to Probability Section 7.3, 7.4, 7.5

Introduction to Probability Section 7.3, 7.4, 7.5 presented by E. G. Gascon

Introduction to Probability Section 7.3 Vocabulary & Examples

Sample Space The set of ALL possible outcomes for an experiment is the sample space. (This will make up the denominator of a probability) Probability range is 0 ≤ P(E) ≤ 1

Event An event is a subset of a sample space defined by the problem (the numerator) Complement of an Event • Is the set of ALL outcomes in a sample space that are not included in the event. Denoted by E’ Example: If the event is throwing an 6 on a dice, then the complement is throwing 1,2,3,4,or 5. So… P(less than 6) = 1 – P(6) This is sometimes an easier way to find a difficult probability

Conditional Probability The Probability of an event occurring, given that another event has already occurred. Denoted by P(B|A), read Probability of B given A. Ex P( Jack of Diamonds) P(Jack | Diamonds) There 1 jack of diamonds P(Jack of Diamonds) = 1/52

IndependentEvents Two event are independent if the occurrence of one of the events does NOT affect the probability of the occurrence of the other Use Conditional probability to determine P(B|A) = P(B) or P(A|B) = P(A) then they are independent Example of Independent Events: Draw two cards WITH replacement. If the first card is a heart, and one puts it back in the deck, then if the second card drawn is red. The probability of drawing a red card does not change because a heart was drawn first. Example of Dependent Events: Draw two cards WITHOUT replacement. If the first card is a heart, then if the second card drawn is red. The probability of drawing a red card changes because a heart was drawn first, therefore there is one less card in the deck, and that heart might have been red.

Mutually Exclusive Events Events E and F are mutually exclusive events if E  F = , meaning that they have no elements in common. • One of these statements will be true: • Event A and B cannot occur at the same time • Event A and B have zero outcome in common • P(A AND B) = 0

Basic Probability Principle Let S be a sample space of equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs is Event note: 0  P(E)  1 Sample space

Intersection Rule for Probability (Multiplication Rule) Two events A AND B occur in sequence P(A and B) = P(A) * P(B|A) SPECIAL CASE: If the events are INDEPENDENT P(A and B) = P(A) * P(B) Take an extra minute to study this slide It is important when faced with an AND problem to decide if the two events are independent (slide # 6).

How to Solve an AND problem Identify the events A and B in sequence Decide whether the events are independent or dependent Find P(A), P(B), or if necessary P(B|A) Use the appropriate Intersection rule (Multiplication rule)

Example AND Problem Two cards are selected at random “without replacement,” find the probability that a King , AND then a Queen is selected. The events are a King then a Queen (Dependent Events) P(K) = P(Q |K) = *

Example Special Case A coin is tossed and a die is rolled Find the probability of getting a head and then rolling a 6 The events are a Heads then a “2” (Independent Events) P(H) = P(“2”) = *

OR For Mathematical purposes OR means INCLUSIVE “OR” Three ways for event A OR B to occur A occurs and B does not B occurs and A does not A and B both occur Ex: In a group, the number wearing Red shirts OR Green Pants will be : # of Red shirts with not green pants + # of green pants with not red shirts + number of Red shirts and Green pants.

Union Rule for Probability (Addition Rule) For ANY events E OR F from a sample space S P(E  F) = P(E) + P(F) – P(E  F) Special Case: For mutually exclusive events E AND F from a sample space S, P(E  F) = P(E) + P(F) (as in examples of section 7.3) Take an extra minute to study this slide It is important when faced with an OR problem to decide if the two events are mutually exclusive. (slide # 7)

How to Solve an OR problem Identify the events A and B Decide whether the events are mutually exclusive Find P(A), P(B), and if necessary P(A and B) Use the appropriate Union rule (ADDITION rule)

Example: Special CaseProbability of drawing a spade OR heart. Probability of drawing a spade. Probability of drawing a Heart. Probability of drawing a spade OR heart. Note: These are mutually Exclusive events.

Example Probability of drawing a red face card. Probability of drawing a red card. There are 26 red cards Probability of drawing a face card. There are 12 face cards BUT, these two events are NOT mutually exclusive... so… there is an extra step Probability of drawing a red face card. There are 6 red –face cards

Solution to Events NOT mutually exclusive Probability of drawing a red OR face card. Note: The difference between this problem and the example on slide 13 On the previous slide #13 was a problem that were mutually exclusive, and this one they were not.

Complement Rule IF you know the probability that event E occurs, then the probability that event E does NOT occur is: P(E’) =1 − P(E) Use this rule to save steps. To find the probability of the roll of two fair dice yields a sum > 3, one could find the P(4) OR P(5) OR P(6) OR P(7) OR P(8) OR P(9) OR P(10) OR P(11) OR P(12) BUT it is easier Find the P( sum 3) = P(1) OR P(2) OR P(3) = 0 + (1/36) + (2/36) = (3/36) = 1/12 so… P( sum >3) = 1 − P(sum  3) = 1 − (1/12) = 11/12

Odds The odds in favor of an event E are defined as the ratio of P(E) to P(E’) The odds that the roll of two fair dice yields a sum > 3 is… P(sum >3) : P(sum  3) =

Conditional Probability Sometimes it is not easy to logically figure the conditional probability, so we have a formula Notice that this formula is a derivation from the AND rule. The conditional probability of event E given event F: Also it can be stated that by re-writing the equation:

Example 7.5 # 17The Problem • Numbers 1,2,3,4,and 5 written on slips of paper, and 2 slips are drawn at random one at a time WITHOUT replacement. • Find the probability that the sum is 8, given the first number is 5. Sample space = {1,2,3,4,5}

Solution • Each number as a 1/5 chance of being drawn for the first slip of paper . • Each number that is left has a ¼ chance of being drawn for the second slip of paper. • There is only 1 way to get a “5” AND “sum of 8”, (5 + 3). • therefore…

Solutions cont. Putting it all together:

Visual Solutions Second Slip drawn First Slip drawn 2 3 4 5 1 3 4 5 1 1 2 2 This is the only scenario which has a sum of 8, and there is only one way to get it, therefore the probability of a sum of 8 given 5 on the first draw is 1/4 4 5 1 3 2 3 4 5 1 2 5 3 4

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Introduction to Probability Section 7.3, 7.4, 7.5