1. Warm-Up A woman and a man (unrelated) each have two children .
At least one of the woman’s children is a boy,
and the man’s older child is a boy.
Do the chances that the woman has two boys equal the chances that the man has two boys?

2. Probability

3. =all possible outcomes
=a possible result
Sample Space
=all possible outcomes
What is the sample space for rolling a single die?
What is the sample space for flipping a coin?
What is the sample space for flipping two coins?

4. Ex. 1
Two coins are tossed. What is the probability that both land heads?
Probability
# of ways both land on heads
All possible outcomes
You can express a probability as a fraction, a decimal, or a percent. For example: , 0.25, or 25%. 1 4
For this problem drawing the sample space is helpful. However, doing so is not always feasible.
What is the probability that one coin lands on heads and one lands on tails?

5. Only one outcome corresponds to rolling a 4.
Ex. 2
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. 1 6 =
number of ways to roll a 4
P (rolling a 4) =
number of ways to roll the die

6. Ex. 2b
You roll a six-sided die whose sides are numbered from 1 through 6. What is the probability of rolling an ODD number?
Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5.
number of ways to roll an odd number 3 6 1 2
= =
P (rolling odd number) =
number of ways to roll the die

7. Ex. 2c
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?
All six outcomes correspond to rolling a number less than 7.
number of ways to roll less than 7
P (rolling less than 7 ) =
number of ways to roll the die 6 =
= 1

8. Impossible and Certain
If P(E) = 0, then the event cannot occur
It is impossible
If P(E) = 1, then the event must occur.
It is certain

9. We could draw out the sample space…
Ex. 3
Two six-sided dice are tossed. What is the probability that the sum of the two dice is 7?
We could draw out the sample space…
1, 1
2, 1
3, 1
4, 1
5, 1
6, 1
1, 2
2, 2
3, 2
4, 2
5, 2
6, 2
1, 3
2, 3
3, 3
4, 3
5, 3
6, 3
1, 4
2, 4
3, 4
4, 4
5, 4
6, 4
1, 5
2, 5
3, 5
4, 5
5, 5
6, 5
1, 6
2, 6
3, 6
4, 6
5, 6
6, 6
But that’s time consuming and tedious…

10. How many ways can I be successful?
Ex. 3
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 2 3 4 5 6
6 total successes 6 5 4 3 2 1

11. Ex. 4
What is the probability of drawing an ace out of a standard deck of cards?
Success:
Drawing an ace:
4 chances
Number of cards:
52 cards
Total:

12. 13C2 = 52C2 number of ways to draw 2 hearts
Ex. 5
Two cards are drawn at random from a standard deck of 52 cards. What is the probability that both are hearts?
number of ways to draw 2 hearts
P(both are hearts)=
number of ways to draw 2 cards
13C2 =
52C2
Can also say (13/52) *(12/51) = 1/17.
If it was draw a heart (replace) and then draw another heart…
(13/52)*(13/52) = 1/16

13. played first, in any order?
Ex. 6
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:
4 C 2
8 C 2 6 28 3 14
P(playing 2 favorites first) = = = 

14. Independent Events Page 758
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

15. 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.
= 6 ways
Sum of 7:
1,6
2,5
3,4
4,3
5,2
6,1
P(sum of 7)
Sum of 4:
1,3
2,2
3,1
= 3 ways
P(sum of 4)
P(sum of 7 and sum of 4)

16. There are 9 brown boxes and 6 red boxes on a shelf.
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)
P(Brian chooses brown)
P(Amanda and Brian choose brown)

17. Mutually Exclusive Events
Any thoughts on what mutually exclusive means?
Ever hear it in conversation?
Page 756

18. Mutually Exclusive Events
…two events that cannot happen at the same time.
…no outcomes in common.
Example: Choosing a spade or a heart.
Example: Tossing a 3 or a 4 on a die.
P(A or B) = P(A) + P(B)
“Or” means add.

19. 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)

20. 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
P(sum of 7)
Sum of 9:
3, 6
4, 5
5, 4
6, 3
P(sum of 9)
P(sum of 7 or sum of 9)

21. Inclusive Events …two events that can happen simultaneously.
Example: Choosing a spade or a two.
Example: Tossing a 4 or a multiple of 2 on a die.
P(A or B) = P(A) + P(B) – P(A and B)
“Or” means add.

22. Ex. 11
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) =
P(face) =
P(heart and face) =

23. 22 ways A 2 3 4 K K K  5 6 Q Q Q  K  7 8 Q  J J J 
Ex. 11
22 ways
A 2
3 4
5 6
7 8
9 10
K K K 
Q Q Q 
J J J 
K 
Q 
J 

24. 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,
find the probability that the student will be able to answer all ten questions on the exam.
# of ways 10 questions can be chosen from 15 known
P(all ten) =
# of ways 10 questions can be chosen from 20 given
15C10
3003 21 = = = ≈
0.016
20C10
184756
1292

25. 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,
find the probability that the student will be able to answer exactly eight questions.
Choose 8 questions from 15 known & 2 from 5 unknown
P(exactly 8) =
Choose 10 from 20 total
15C8
• 5C2
6435 • 10
225
0.348 = = = ≈
20C10
184756
646

26. • 5C1 • 5C0 P(at least 9) = P(9 or 10) = P(9) + P(10) 15C9 15C10 = +
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,
find the probability that the student will be able to answer at least nine questions.
P(at least 9) =
P(9 or 10) =
P(9) + P(10)
15C9
• 5C1
15C10
• 5C0 = +
20C10
20C10
5005 • 5
3003 • 1 = +
184756
184756
28028 49 = = 
184756
323

27. At least 3 girls means that 3 or 4 girls could be chosen.
Ex. 13
There are 6 children in the Patel family, 4 girls and 2 boys. Four children are chosen to model for a fashion show. What is the probability that at least three girls are chosen.
At least 3 girls means that 3 or 4 girls could be chosen.
P(at least 3 girls)
= P(3 girls) + P(4 girls)
P(3 girls) = P(3 girls and 1 boy)
P(4 girls) = P(4 girls and 0 boys)
P(at least 3 girls)
= P(3 girls) + P(4 girls)

28. An event A and its complement A’ must sum to 1.
Complement of an Event
Winning a game
Losing a game
Raining
Not Raining
Walking to school
Not walking to school
An event A and its complement A’ must sum to 1.
P(A’) = 1 – P(A)
P(A) = 1 – P(A’)

29. A bag contains four yellow, and three red marbles.
Ex. 14
A bag contains four yellow, and three red marbles.
Two marbles are chosen (without replacement).
Find the probability that the marbles are different.
What’s the probability that the marbles will be the same color?
So the probability that the marbles will be different colors...

30. Two-Children Problem Solution:
The chances that the woman has two boys are1 in 3
and the chances that the man has two boys are 1 in 2.

31. Sample space: BB, BG, GB, GG
Two-Children Problem
Sample space: BB, BG, GB, GG
For the man the sample space reduces to: BB and BG.
Hence, the probability that he has two boys is 1 out of 2.
For the woman the sample space reduces to: BB, BG and GB.
Thus, her chances of having two boys is 1 out of 3.

32. Homework Pg 682 More Probability Practice Worksheet
35, 37, 42, 45, 46, 51, 56, 57
More Probability Practice Worksheet
WS Review §9.6, §9.7
51 is tricky, 56 is tricky