probability we love section 9.3a and b!. most people have an intuitive sense of probability, but...

ProbabilityProbabilityWe love Section 9.3a and b!We love Section 9.3a and b!

Most people have an Most people have an intuitive senseintuitive sense of ofprobability, but that intuition is oftenprobability, but that intuition is oftenincorrect…incorrect…

Let’s test your intuitionLet’s test your intuitionabout probability!!!about probability!!!

Intuition about Probability?Intuition about Probability?

Find the probability of each of the following events.

1. Tossing a head on one toss of a fair coin.

Two equally likely outcomes: { T, H }. Probability is 1/2.Two equally likely outcomes: { T, H }. Probability is 1/2.

2. Tossing two heads in a row on two tosses of a fair coin.

Four equally likely outcomes: { TT, TH, HT, HH }.Four equally likely outcomes: { TT, TH, HT, HH }.Probability is 1/4.Probability is 1/4.



3. Drawing a queen from a standard deck of 52 cards.

There are 52 equally likely outcomes, 4 of which areThere are 52 equally likely outcomes, 4 of which arequeens. Probability is 4/52, or 1/13.queens. Probability is 4/52, or 1/13.

4. Rolling a sum of 4 on a single roll of two fair dice.

Multiplication Principle of Counting Multiplication Principle of Counting 6 x 6 = 36 equally 6 x 6 = 36 equallylikely outcomes. Of these, three { (1, 3), (3, 1), (2, 2) }likely outcomes. Of these, three { (1, 3), (3, 1), (2, 2) }yield a sum of 4. Probability is 3/36, or 1/12.yield a sum of 4. Probability is 3/36, or 1/12.



5. Guessing all 6 numbers in a state lottery that requiresyou to pick 6 numbers between 1 and 46, inclusive.

Number of equally likely ways to choose 6 numbersNumber of equally likely ways to choose 6 numbersfrom 46 numbers without regard to order?from 46 numbers without regard to order?

46 6 9,366,819C Probability is 1/9366819.Probability is 1/9366819.

Terminology of ProbabilityTerminology of Probability

Sample Space – the set of all possible outcomesof an experiment

Event – a subset of the sample space

Each sample space consists of a finite number ofequally likely outcomes…

Probability of an EventProbability of an Event(Equally Likely Outcomes)(Equally Likely Outcomes)

If E is an event in a finite, nonempty sample spaceS of equally likely outcomes, then the probabilityof the event E is

P(E) =the number of outcomes in E

the number of outcomes in S

A key part of the hypothesis!!!A key part of the hypothesis!!!

Rolling DiceRolling Dice

Possible outcomes for the sum on two fair dice:

{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }

So the probability of rolling a sum of 4 is 1/11, right???

NO!!!, because these sums are not equally likely…

Possible outcomes when rolling a BLUEBLUE and Aqua Aqua die:

11--11 Sum = 2

11--22, , 22--11 Sum = 3

11--33, , 22--22, , 33--11 Sum = 4

11--44, , 22--33, , 33--22, , 44--11 Sum = 5

11--55, , 22--44, , 33--33, , 44--22, , 55--11 Sum = 6

11--66, , 22--55, , 33--44, , 44--33, , 55--22, , 66--11 Sum = 7

22--66, , 33--55, , 44--44, , 55--33, , 66--22 Sum = 8

33--66, , 44--55, , 55--44, , 66--33 Sum = 9

44--66, , 55--55, , 66--44 Sum = 10

55--66, , 66--55 Sum = 11

66--66 Sum = 12

Probability DistributionProbability Distribution

Outcome Probability

2 1/36

3 2/36

4 3/36

5 4/36

6 5/36

7 6/36

8 5/36

9 4/36

10 3/36

11 2/36

12 1/36Note: The sum of the probabilities ofNote: The sum of the probabilities of

any prob. dist. is always 1!!!any prob. dist. is always 1!!!

Definition: Probability FunctionDefinition: Probability Function

A probability function is a function P that assigns a realnumber to each outcome in a sample space S subject tothe following conditions:

1. 0 1P O for every outcome O.

2. The sum of the probabilities of all outcomes in S is 1.

3. 0P The The empty setempty set……

Probability of an EventProbability of an Event(Outcomes not Equally Likely)(Outcomes not Equally Likely)

Let S be a finite, nonempty sample space in whichevery outcome has a probability assigned to it bya probability function P. If E is any event in S, theprobability of the event E is the sum of theprobabilities of all the outcomes in E.

Returning to our dice…Returning to our dice…

Suppose that two fair dice have been rolled. State theprobability of each of the following events.

1. The sum is 7. P(sum 7) = 1/6P(sum 7) = 1/6

2. The same number is rolled on both dice.P(doubles) = 1/6P(doubles) = 1/6

3. The sum is 2 or 3. P(sum 2 or 3) = 1/12P(sum 2 or 3) = 1/12

4. The sum is a multiple of 3. P(multiple of 3) = 1/3P(multiple of 3) = 1/3

Strategy for Determining ProbabilitiesStrategy for Determining Probabilities1. Determine the sample space of all possible outcomes. When possible, choose outcomes that are equally likely.

2. If the sample space has equally likely outcomes, the

Probability of an event E is determined by counting:

P( E ) =the number of outcomes in E

the number of outcomes in S3. If the sample space does not have equally likely outcomes, determine the probability function. (This is not always easy to do.) Check to be sure that the conditions of a probability function are satisfied. Then the probability of an event E is determined by adding up the probabilities of all the

outcomes contained in E.

Multiplication Principle of Probability

Suppose an event A has probability p and an event Bhas probability p under the assumption that A occurs.then the probability that both A and B occur is p p .

1

2

1 2

If the events A and B are independent, we omit thephrase “under the assumption that A occurs.”

Practice ProblemsPractice Problems

Sal opens a box of a dozen chocolate covered doughnuts andoffers two of them to Val. Val likes the vanilla-filled doughnutsthe best, but all of the doughnuts look alike on the outside. Iffour of the twelve doughnuts are vanilla-filled, what is theprobability that both of Val’s picks turn out to be vanilla?

Here, we are choosing two doughnuts (without regard to order)from a box of 12:

12 2 66C outcomes of this experiment

Are they all equally likely? YES!!!


Sal opens a box of a dozen chocolate covered doughnuts andoffers two of them to Val. Val likes the vanilla-filled doughnutsthe best, but all of the doughnuts look alike on the outside. Iffour of the twelve doughnuts are vanilla-filled, what is theprobability that both of Val’s picks turn out to be vanilla?

The event E consists of all possible pairs of 2 vanilla-filleddoughnuts (without regard to order) from the 4 vanilla-filleddoughnuts available:

4 2 6C ways to form such pairs

6 66 1 11P E


Another way to solve the same problem:

We want the probability of outcome VV.

There are two types of doughnuts: vanilla (V) and unvanilla (U)

When choosing two doughnuts, there are four possibleoutcomes: { VV, VU, UV, UU }

Are they all equally likely? NO WAY!!!


There are two types of doughnuts: vanilla (V) and unvanilla (U)

When choosing two doughnuts, there are four possibleoutcomes: { VV, VU, UV, UU }

Probability of picking a V on the first draw:

4/12

Probability of picking a V on the second draw, under theassumption that a V was drawn on the first:

3/11

Probability of drawing a V on both draws(use the Mult Prop!!!):

4 3 1

12 11 11

Practice ProblemsPractice ProblemsNow, I want you to create a probability function for this experiment…

Outcome Probability

V V 1/11

V U (4/12)(8/11) = 8/33

U V (8/12)(4/11) = 8/33

U U (8/12)(7/11) = 14/33

Does this probability function “check out?”

Practice ProblemsPractice ProblemsSkittles candy comes in the following color proportions:

Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3

Purple

0.1

A single Skittle is selected at random from a newly-opened bag.What is the probability that the candy has the given color(s)?

1. Green or Orange P(G or O) = P(G) + P(O) = 0.2 + 0.3 = 0.5P(G or O) = P(G) + P(O) = 0.2 + 0.3 = 0.5

2. Purple or Orange or Red

P(P or O or R) = P(P) + P(O) + P(R)P(P or O or R) = P(P) + P(O) + P(R) = 0.1 + 0.3 + 0.25 = 0.65= 0.1 + 0.3 + 0.25 = 0.65


Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3

Purple

0.1

A single Skittle is selected at random from a newly-opened bag.What is the probability that the candy has the given color(s)?

3. Not Yellow P(not Y) = 1 – P(Y) = 1 – 0.15 = 0.85P(not Y) = 1 – P(Y) = 1 – 0.15 = 0.85

4. Neither Green nor Red

P[not (G or R)] = 1 – P(G or R) = 1 – (0.2 + 0.25)P[not (G or R)] = 1 – P(G or R) = 1 – (0.2 + 0.25) = 0.55= 0.55


Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3Purple

0.1

Now, a Skittle is selected at random from each of three newly-opened bags. What is the probability that the three Skittleshave the given color(s)?

5. All three are Orange

P(O1 and O2 and O3) = P(O1) x P(O2) x P(O3)P(O1 and O2 and O3) = P(O1) x P(O2) x P(O3)

= (0.3)(0.3)(0.3) = 0.027= (0.3)(0.3)(0.3) = 0.027


Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3

Purple

0.1


6. All three are Purple

P(P1 and P2 and P3) = P(P1) x P(P2) x P(P3)P(P1 and P2 and P3) = P(P1) x P(P2) x P(P3)

= (0.1)(0.1)(0.1) = 0.001= (0.1)(0.1)(0.1) = 0.001


Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3

Purple

0.1


7. None are Red

P(none R) = P(not R1 and not R2 and not R3)P(none R) = P(not R1 and not R2 and not R3)

= P(not R1) x P(not R2) x P(not R3)= P(not R1) x P(not R2) x P(not R3)

= (0.75)(0.75)(0.75) = 0.421875= (0.75)(0.75)(0.75) = 0.421875


Color

Proportion

Red

0.25

Green

0.2

Yellow

0.15

Orange

0.3

Purple

0.1


8. The 1st is Red, the 2nd is Orange, and the 3rd is not Purple

P(R1 and O2 and not P3) = P(R1) x P(O2) x P(not P3)P(R1 and O2 and not P3) = P(R1) x P(O2) x P(not P3)

= (0.25)(0.3)(0.9) = 0.0675= (0.25)(0.3)(0.9) = 0.0675

probability we love section 9.3a and b!. most people have an intuitive sense of probability, but...

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