04.02 Probability

Practice 3 Suppose 60% of homes in Miami have a swimming pool and 30% have both a swimming pool and a Jacuzzi. What is the probability that a randomly selected home will have a Jacuzzi given that it has a swimming pool?

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Example 2 You and a friend are going shopping for a new car. At the car dealership, you can choose from an electric car, a gasoline car, and a hybrid vehicle. You want to choose between electric and hybrid, and your friend wants to purchase only an electric vehicle. Describe this scenario using sets. Let E equal {electric, gasoline, hybrid} Let C equal {electric, hybrid} Let D equal {electric}. 1. What is C ∩ D? (intersection) 2. What is C U D? (union) 3. What is D'? (complement)

1. C ∩ D equal {electric}. This is the only element that is in both sets C and D. 2. C U D equal {electric, hybrid}. These are the elements that are in either set C or set D. 3. D' equal {gasoline, hybrid}. These are the elements in set E that are not in set D. Take a closer look at sets D and C. Did you notice that set D is included in set C? This can be described using set notation by using the symbol ⊆. Write D ⊆ C to show that all elements in set D are also in set C.

Hillary rolls two dice numbered 1 through 6 while playing her favorite board game. She will get a second turn if she rolls a sum that is an odd number greater than 10. What are Hillary's chances of getting a second turn when she rolls the dice?

1/18

Practice 2 You roll two six-sided dice. What is the probability that the sum will be eight given that one die shows a 5?

1/6

Intersection

The intersection of two sets is the set that contains elements that are common to both sets. Intersections are written mathematically as A ∩ B.

Example 8 At Florida State University, 80% of students take a math course. Seventy percent of those who don't take a math course take English 101, whereas only 50% of those who do take a math course take English 101. What is the probability that a student takes English 101? What is the probability that a student who takes English 101 doesn't take a math course?

Calculate across to find the probabilities: P(English 101 | Math) = (0.8)(0.5) = 0.4 P(English 101 | No Math) = (0.2)(0.7) = 0.14. Then, add the probabilities for all students who take English 101: P(student takes English 101) = 0.4 + 0.14 = 0.54. A student at Florida State University, chosen at random, has a 0.54 probability of taking English 101. P(No math | English 101) = P(no math and takes English 101)/P(English 101) = 0.14/(0.4+0.14) = 0.14/0.54 = 0.2593 So the probability that a student who takes English 101 will not take a math course is approximately 0.26.

Example 10 Dwayne Wade has a 0.7 probability of making a free throw. However, if he makes the first shot, the probability of making the second shot drops to 0.5. What is the probability of him making two consecutive free throws?

Dwayne Wade's two free throws are not independent events because the outcome of the first throw affects the probability of the second. We must use the formula P(A ∩ B) = P(A) • P(B | A). P(Making both free throws) = P(Making the first free throw) × P(Making the second free throw given he makes the first) = 0.7 • 0.5 = 0.35 The probability of Dwayne Wade making both free throws is 35%. Notice, this is not the same thing as (0.7)(0.7)=0.49! This is a common error that students make on this kind of problem.

Example 9 A coin is tossed twice. What is the probability that it will land with "heads" showing both times?

Flipping a coin the second time is independent from the first flip—meaning, the outcome of the first flip does not affect the second flip. P(heads∩heads) = P(heads)•P(heads) = (1/2)(1/2) = 1/4 = 0.25 = 25% The probability that the coin will land with heads showing both times is equal to 25%.

Independent events: multiplication rule

If events A and B are independent, then P(A and B) = P(A∩B) = P(A)P(B). That is, if you have independence, you can simply multiply the probabilities of A and B to get the probability of A and B.

Example 12 A string of holiday lights contains 25 bulbs. Each of the bulbs functions independently. If each bulb has a 0.015 chance of failing over a three-year period, what is the probability that the entire string will function after three years without maintenance?

If the chance of the sting failing is 0.015, then the complement represents the chance of them not failing, which is 1 - 0.015 = 0.985. Because each light works independently, you can multiply the probabilities. This means P(First light works and second light works and third light works and fourth light works . . . and 25th light works), which is just 0.985 multiplied by itself 25 times, or 0.985^25, which equals 0.6853. There is a 68.53% chance that the string of lights will still function after three years.

Example 11 Event A: choosing a jack P(A) = 4/52 Event B: choosing a spade P(B) = 13/52

If we simply added P(A) + P(B) to get 452+1352, we would count the jack of spades twice. These two events are not disjoint, because the jack of spades is an outcome in common. Therefore, use the general addition rule. P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 − 1/52 = 1652 or 0.3077 The probability of getting a jack and a spade is 1 out of 52. Only one of these jacks exists in the entire deck.

Example 7 In the Tour de France, all cyclists have to go through mandatory drug testing. Ephedrine (available over the counter as Sudafed) is one of the banned drugs tested. Here are a few questions we might ask: -What percentage of cyclists are taking ephedrine and getting away with it? -Of the cyclists who are taking ephedrine, what percentage might we expect to catch? -What percentage of cyclists who test positive for ephedrine are really innocent? -For what percentage of nonephedrine-doping cyclists will the test work correctly? -If we select a cyclist at random,What is the chance that he or she is taking ephedrine? -What is the chance that the test will conclude he or she is taking ephedrine?

Information known: -Five percent of cyclists use ephedrine. -From prior research, we know that 1% of cyclists use ephedrine and get away with it. -If ephedrine is not in the body's system, the test will come out clear 99.3% of the time. Organize and sort the information into a tree diagram. Fill in the probabilities you are given: -If 0.05 cyclists use ephedrine, 0.95 don't. -If 0.01 of users test negative, then 0.99 test positive. -If 0.993 of nonusers test negative, then 0.007 nonusers test positive. To find the remaining probabilities, multiply along the tree's branches.

Example 6 A die is rolled. The probability of rolling a 5 given that an odd number is rolled is 1/3. The probability of rolling an odd number is 1/2. Using the probability formula, what is the probability of rolling a 5 and rolling an odd number?

Let A equal rolling a 5 and B equal rolling an odd number. Begin with the formula: P(A | B) = P(A∩B)/P(B) Substitute what you know: 1/3 = P(A∩B)/(1/2) Then, solve for P(A∩B): 1/3 = P(A∩B)/(1/2) (0.5)(1/3) = P(A∩B) 1/6 = P(A∩B)

Example 13 Government data show that 26% of the civilian labor force has at least four years of college and 16% of that labor force works as laborers or operators of machines or vehicles. Can you conclude that because (0.26)(0.16) = 0.0416, about 4% of the civilian labor force includes college-educated laborers or operators?

No. To multiply probabilities, they must be independent. It is very possible that whether you have a college education affects whether you are a laborer/operator. These events are not independent.

Practice 4 Given P(A) = 0.3, P(B) = 0.4, and P(A|B) = 0.3, what is P(A and B) and P(A or B)?

P(A and B) = 0.12, P(A or B) = 0.58

Dependent events: multiplication rule

P(A ∩ B) = P(A) • P(B | A)

Addition Rule

P(A∪B) = P(A)+P(B)−P(A∩B) The formula for probability addition is included in your AP Resource Packet.

In the case of events A and B being mutually exclusive, P(A∩B)=0, so P(A∪B)=P(A)+P(B)−P(A∩B)=P(A)+P(B)−0=

P(A∪B)= P(A)+P(B)−P(A∩B) = P(A)+P(B)−0 = P(A)+P(B). For example, event A is choosing a face card and event B is choosing a 5. There is no way to choose a face card that is also a 5. These are disjoint, or mutually exclusive. Event A is choosing a 5 and event B is choosing a spade. Because there is a 5 of spades (an outcome in A and B), these events are not disjoint, or mutually exclusive.

Practice 1 The 2010 US Census identified the approximate ethnic breakdown for the state of Florida to be white, 78%; black, 14%; native American and native Alaskan, 1%; Asian, 2%; native Hawaiian and other Pacific islander, 1%; other, 3%; two or more races, 1%. Assuming these are mutually exclusive categories, what is the probability that a randomly selected person from the state of Florida will be black or Asian?

P(black or Asian) = P(black) + P(Asian) = P(0.16) + P(0.02) = 0.16 or 16%

General addition rule

The General Addition Rule applies to all events: P(A or B) = P(A) + P(B) − P(A and B) P(A∪B) = P(A) + P(B) − P(A∩B)

Complement

The complement of a set contains all elements in the universal set that are not in that particular set. The complement of set A is written as A'.

Complement Rule

The complement of any event A is the probability that event A will not occur. If the probability that it will snow today is 40%, then the probability that it will not snow is 60%. S = It will snow. Then, S^c = it will not snow. P(S) = 0.4 P(S^c) = 1−P(S) = 1−0.4 = 0.6

Example 3 What is the probability of choosing a 5 or a face card from a standard deck of cards?

The probability of choosing a 5 is 4/52 because there are four 5s in a standard deck of 52 cards. The probability of choosing a face card is 12/52 because there are 12 face cards (Jack, Queen, King in each of four suits) in the deck of 52 cards. Because they are mutually exclusive, the probability is P(5 or face card) = 452+1252 = 1652 or 0.3077. The probability of choosing a 5 or a face card from a standard deck of cards is 0.3077.

Union

The union of two sets is the set that contains all elements of those two sets. In other words, all numbers within both circles in the Venn diagram. The union is written mathematically as A U B.

Example 5 A card is chosen at random from a deck. What is the probability you will pick a Jack given that the card chosen is a face card?

There are 52 cards in a standard deck and four suits (hearts, spades, diamonds, and clubs). Each suit has three face cards, so there are 12 face cards in total. The probability of choosing a face card is 12/52. Each suit has one Jack. The probability of choosing a jack is 452. P(Jack | face card) =P(Jack and face card) / P(face card)=(4/52)/(12/52) = 4/12 = 13 = 33%. So, given that you chose a face card from a standard deck, there is a 33% probability the card will be a Jack.

Example 4 A die is rolled. What is the probability that you will roll a 6 given that you rolled an even number?

There are six sides on a die, and only one side has the number 6. So, the probability of rolling a 6 is 1/6. There are three sides on a die with even numbers (2, 4, and 6). So, the probability of rolling an even number is 3/6. When you put it together in the formula, you get: P(6 | even) = P(6 and even) / P(even) = (1/6)/(3/6) = 1/3 = 33.33%. This probability tells you that, assuming you roll an even number, you will have a 33.33% chance of that even number being a 6.

Set Notation

Universal Set Intersection Union Complement

Example 1 Take a look at the Venn diagram about electives, and label some sets and subsets. Let set E equal the universal set (all students in the school). Let set A equal music students. Let set B equal drama students.

Universal set: 990 (all students in school) Intersection: The intersection of sets A and B is the students who are taking both music and drama—in other words, the number of students in the overlapping portion of the Venn diagram. There are 342 students in A ∩ B. Union: In the diagram above, A U B includes 900 students—the 265 music students, the 293 drama students, and the 342 students taking both music and drama. Notice that this set does not include the 90 students who do not want to take music or drama. If those students were included, you would be looking at the universal set. Complement: The complement of set A is all students in set E who are not taking music, which would be 383 students (990 − 265 − 342 = 383). The complement of set A is written as A'.

Why use conditional probability?

We use conditional probability to prove whether two events are independent—meaning, the outcome of one event does not influence the probability of the other. If P(A|B) = P(A), then we can say events A and B are independent.

Disjoint

When two events are mutually exclusive; they have no outcomes in common For example, the probability of being both a junior and a senior in high school is zero because it is not possible to be both at once.

Conditional probability

When we want to find the probability of an event given another event has occurred; written as P(A|B), which reads "the probability of A given B". We can use a formula to determine conditional probability.

The probability P(A) of any event is always...

a number between zero and one.

P(A|B) = P(A∩B) / P(B)

conditional probability formula; emember that whatever event you are "given" should be in the denominator!

What is the probability of choosing a 2 or a 10 from a standard deck of cards? Answer as a fraction (blank)/ (blank)

fractional answer 2/13

Universal Set

the set that contains all elements of the sample space

Tree diagram

to organize possible outcomes into increasingly specific combinations of outcomes; goal is to have an accurate list of all events in the sample space to use in calculating probabilities