Warm-Up A woman and a man (unrelated) each have A woman and a man (unrelated) each have

two children . two children . At least one of the woman’s children is a boy, At least one of the woman’s children is a boy, and the man’s older child is a boy. and the man’s older child is a boy. Do the chances that the woman has two Do the chances that the woman has two

boys equal the chances that the man has two boys equal the chances that the man has two boys?boys?

Probability

Sample Space =all possible outcomes

Outcome =a possible result

What is the sample space for rolling a single die?

}6,5,4,3,2,1{S

What is the sample space for flipping a coin?

},{ THS

What is the sample space for flipping two coins? },,,{ TTTHHTHHS

Two coins are tossed. What is the probability that both land heads?

All possible outcomes

# of ways both land on heads

Ex. 1

},,,{ TTTHHTHHS

P HH( ) 1

4

Total

Successobability PrProbability

You can express a probability as a fraction, a decimal, or a percent.For example: , 0.25, or 25%.1

4

You roll a six-sided die whose sides are numbered from 1 through 6. What is the probability of rolling a 4?

Only one outcome corresponds to rolling a 4.

P (rolling a 4) = number of ways to roll a 4number of ways to roll the die

16

=

Ex. 2

Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5.

P (rolling odd number) = number of ways to roll an odd number

number of ways to roll the die

You roll a six-sided die whose sides are numbered from 1 through 6. What is the probability of rolling an ODD number?

36

12

= =

Ex. 2b

All six outcomes correspond to rolling a number less than 7.

P (rolling less than 7 ) = number of ways to roll less than 7

You roll a six-sided die whose sides are numbered from 1 through 6. What is the probability of rolling a number less than 7?

number of ways to roll the die

66= = 1

Ex. 2c

If P(E) = 0, then the event cannot occur

Impossible and Impossible and CertainCertain

If P(E) = 1, then the event must occur.

It is impossible

It is certain

10 yprobabilit

Two six-sided dice are tossed. What is the probability that the sum of the two dice is

7?

Ex. 3

We could draw out the sample space…

1, 2

1, 3

1, 4

1, 1

1, 5

1, 6

2, 2

2, 3

2, 4

2, 1

2, 5

2, 6

3, 2

3, 3

3, 4

3, 1

3, 5

3, 6

4, 2

4, 3

4, 4

4, 1

4, 5

4, 6

5, 2

5, 3

5, 4

5, 1

5, 5

5, 6

6, 2

6, 3

6, 4

6, 1

6, 5

6, 6

But that’s time consuming and tedious…

Two six-sided dice are tossed. What is the probability that the sum of the two dice is

7?Use the fundamental counting principle to figure out the sample space (total possible

outcomes)6 6 36 total outcomes

How many ways can I be successful?

1

6

2 3 4 5 6

5 4 3 2 16 total successes

Ex. 3

P( )sumof 76

36

1

6

What is the probability of drawing an ace out of a standard deck of cards?

Success: Drawing an ace: 4 chances

Total: Number of cards: 52 cards

Ex. 4

P( )drawing anace 4

52

1

13

Two cards are drawn at random from a standard deck of 52 cards. What is the probability that both are hearts?

13C2

52C2

Ex. 5

78

1326

number of ways to draw 2 hearts

number of ways to draw 2 cards

1

17

P(both are hearts)=

=

You put a CD that has 8 songs in your CD player. You set the player to play the songs at random. The player plays all 8 songs without repeating any song.

You have 4 favorite songs on the CD. What is the probability that 2 of your favorite songs are played first, in any order?

There are 8C2 different combinations of 2 songs. Of these,

4C2 contain 2 of your favorite songs. So, the probability is:

P(playing 2 favorites first) = 4 C 2

8 C 2

628

314

Ex. 6

0.214= =

Independent Events

If two events, A and B, are independent, then the probability of both events occurring is...

P(A and B) = P(A) * P(B)

“and” means multiply

Page 758

Ex. 7 Find the probability of getting sum of 7 on the first toss of two dice and a sum of 4 on the second toss.

P(sum of 7)

Sum of 7: 1,6 2,5 3,4 4,3 5,2 6,1 = 6 ways

36

6

P(sum of 4)

Sum of 4: 1,3 2,2 3,1

36

3

= 3 ways

P(sum of 7 and sum of 4)36

3*

36

6

72

1

Ex. 8 There are 9 brown boxes and 6 red boxes on a shelf. Amanda chooses a box and replaces it. Brian does the same thing. What is the probability they both choose a brown box?

The events are independent because Amanda replaced the box.

P(Amanda chooses brown)15

9

P(Brian chooses brown)15

9

P(Amanda and Brian choose brown)15

9*

15

9

25

9

Mutually Exclusive Events

Any thoughts on what mutually exclusive means?

Ever hear it in conversation?

Page 756

…two events that cannot happen at the same time.

Mutually Exclusive Events

Example: Choosing a spadespade or a heartheart.

Example: Tossing a 33 or a 44 on a die.

P(A or B) = P(A) + P(B)

“Or” means add.

)()()( BPAPBAP

…no outcomes in common.

Ex. 9 Christy has 6 pennies, 4 nickels and 5 dimes in her pocket. She takes one coin from her pocket at random. What is the probability that it is a nickel or a dime?

P(nickel) P(dime)

P(nickel or dime)

15

4

15

5

15

5

15

4

5

3

15

9

Ex. 10 Find the probability of getting sum of 7 or a sum of 9 when tossing 2 dice.

Sum of 7: 1,6 2,5 3,4 4,3 5,2 6,1

Sum of 9: 3, 6 4, 5 5, 4 6, 3

P(sum of 7)36

6

P(sum of 9)

P(sum of 7 or sum of 9)

4

36

6

36

4

36

10

36

5

18

…two events that can happen simultaneously.Example: Choosing a spade spade or a twotwo.

Example: Tossing a 44 or a multiple of 2multiple of 2 on a die.

P(A or B) = P(A) + P(B) – P(A and B)

“Or” means add.

Inclusive Events

)()()()( BAPBPAPBAP

One card is selected from a standard deck of 52 playing cards. What is the probability that the card is either a heart or a face card?

P(heart) =52

13P(face) =

52

12P(heart and face) =

52

3

Ex. 11

P( )heart or face P( )heart P( )face P( )heart and face

P( )heart or face 13

52

12

52

3

52

P( )heart or face 22

52

11

26

A 2

3 4

5 6

7 8

9 10

K K K

Q Q Q

J J J

K

Q

J

22 waysEx. 11

Ex. 12a A class is given a list of 20 study problems from which ten will be a part of an upcoming exam.

If a student knows how to solve 15 of the problems,

P(all ten) = # of ways 10 questions can be chosen from 15 known# of ways 10 questions can be chosen from 20 given

15C10

20C10

3003184756

211292

== ≈ 0.016

find the probability that the student will be able to answer all tenall ten questions on the exam.

=

Ex. 12b A class is given a list of 20 study problems from which ten will be a part of an upcoming exam.

If a student knows how to solve 15 of the problems,

P(exactly 8) =Choose 8 questions from 15 known & 2 from 5 unknown

Choose 10 from 20 total

15C8

20C10

= 6435 • 10184756

225646

== 0.348

find the probability that the student will be able to answer exactly eight exactly eight questions.

• 5C2 ≈

Ex. 12c A class is given a list of 20 study problems from which ten will be a part of an upcoming exam.

If a student knows how to solve 15 of the problems,

P(at least 9) =

15C9

20C10

5005 • 5184756

28028184756

+=

0.1517

find the probability that the student will be able to answer at least nine at least nine questions.

• 5C1

P(9 or 10) = P(9) + P(10)

15C10

20C10

• 5C0

+= 3003 • 1184756

= 49323

=

Winning a game Losing a game

Raining Not Raining

Walking to school Not walking to school

P(A’) = 1 – P(A)

8

3

8

5

An event A and its complement A’ must sum to 1.

P(A) = 1 – P(A’)

Complement of an Event

Ex. 14

A bag contains four yellow, and three red A bag contains four yellow, and three red marbles.marbles.

Two marbles are chosen (without replacement).Two marbles are chosen (without replacement). Find the probability that the marbles are Find the probability that the marbles are

different.different.

What’s the probability that the marbles will be the same color?

27

2324

C

CC

So the probability that the marbles will be different colors...

7

31

7

3

7

4

Two-Children Problem

Solution:Solution: The chances that the woman has two boys The chances that the woman has two boys

are1 in 3 are1 in 3 and the chances that the man has two boys and the chances that the man has two boys

are 1 in 2.are 1 in 2.

Two-Children ProblemSample space:Sample space: BB, BG, GB, GGBB, BG, GB, GG

For the man the sample space reduces to: BB For the man the sample space reduces to: BB and BG.and BG. Hence, the probability that he has two boysHence, the probability that he has two boys is 1 out of 2. is 1 out of 2.

For the woman the sample space reduces to: For the woman the sample space reduces to: BB, BG and GB.BB, BG and GB. Thus, her chances of having two boys is 1 out of 3.Thus, her chances of having two boys is 1 out of 3.

Homework

Pg 682Pg 682 35, 37, 42, 45, 46, 51, 56, 5735, 37, 42, 45, 46, 51, 56, 57

More Probability Practice WorksheetMore Probability Practice Worksheet WS Review §9.6, §9.7WS Review §9.6, §9.7