Transcript Probability

Probability
What are the chances?
Definition of Probability
Probability is the likelihood of an event occur. This event could be randomly
selecting the ace of spades, or randomly selecting a red sock or a thunderstorm.
Every possibility for an event
is called an outcome. For
instance, if the event is
randomly drawing a card,
there are 52 outcomes.
We define probability as
# of ways to have success
# of possible outcomes
This is called the sample space
How
many
ways can I
win?
All probabilities are between 0 and 1. That means there are always more possible
outcomes than successful outcomes.
Counting
To solve basic probability questions, we will
need to find two numbers:
This may involve a lot of counting. Tree
# of ways to have success diagrams and the FUNdamental
Counting Theorem will help.
# of possible outcomes
Ex1: A university student needs to take a language course, a math course and
a science course. There are 2 language courses available (English and French),
3 math courses to choose from (Stats, Calculus and Algebra) and 2 science
courses available (Physics and Geology). How many possible schedules are
there?
In other words, What is the sample space?
Let’s draw a First Course:
tree
diagram to Second Course:
show the
entire
P
Third
Course:
sample
space?
E
S
G
C
P
Or
F
A
G
P
S
G
P
G
C
There are
12 possible
schedules
A
P
P
G
G
Counting
Ex 2. A family has 3 children. What is the probability that the 2
youngest will be boys?
# of ways to have success
# of possible outcomes
1st child
B
2nd child
B G
3rd child
There are 8
possible families
G
B G
B G B G
How many have the
2 youngest as boys?
B G B G
2: # of ways to have
success
P(3 kids, 2 youngest are boys) = 2/8 or 1/4
These tree diagrams are great because they show the entire sample
space. They can be cumbersome, though.
Counting with the FTC
We can see that to count the total possible
outcomes, we look at the outcomes of each stage:
From Ex 1:
____
____
Course 1
Course 2
____
=
Course 3
This works if we multiply
the number of outcomes
at each stage.
The fundamental counting
theorem states: to
calculate the sample space
of a multi-staged event,
multiply the number of
outcomes at each stage.
Remember, if you’re drawing blanks,
draw blanks.
Finding the Sample Space
Ex 3. What is the sample space for each event?
a.
b.
c.
d.
e.
Rolling a die
Flipping 3 coins
Drawing a card
Drawing 2 cards
Drawing 1 card, putting it back,
then drawing another.
Ex 4. I have 3 shirts, 6 pants and 4
pairs of shoes. How many (random)
outfits can I create?
3
x ____
6
____
x ____
4
= 72
a. There are 6 outcomes.
b. ___
2 x ___
2 x ___
2 =8
c. There are 52 outcomes.
d. ___
52 x ___
51 = 2652
e. ___
52 x ___
52 = 2704
And or Or
In Probability, the words ‘and’ and ‘or’ are of huge importance.
‘And’ means that
BOTH events occur.
‘Or’ means that ONE
OF the events occur.
# of ways to have success
Ex. A pair of dice is rolled. What is the
probability of rolling
# of possible outcomes
a. A six on the first
a. To win in this situation I must roll 2
AND a five on the second?
numbers, therefore there are 2 stages (draw
b. A three on the first
blanks)
___
___
AND a three on the second?
P(rolling a 6 and a 5) = _____________
c. An even number on the first AND
an even on the second?
___
___
d. A 3 on the first and a 3 on the second
OR a 1 on the first and a 1 on the
P(rolling a 6 and a 5) = 1/36
second.
And or Or
In Probability, the words ‘and’ and ‘or’ are of huge importance.
‘And’ means that
BOTH events occur.
‘Or’ means that ONE
OF the events occur.
# of ways to have success
Ex. A pair of dice is rolled. What is the
probability of rolling
# of possible outcomes
a. A six on the first
b. To win in this situation I must roll 2
AND a five on the second?
numbers (2 blanks)
b. A three on the first
AND a three on the second?
P(rolling a 3 AND a 3)
_____________
=
c. An even number on the first AND
an even on the second?
d. A 3 on the first and a 3 on the second
OR a 1 on the first and a 1 on the
P(rolling a 3 AND a 3) = 1/36
second.
And or Or
In Probability, the words ‘and’ and ‘or’ are of huge importance.
‘And’ means that
BOTH events occur.
‘Or’ means that ONE
OF the events occur.
# of ways to have success
Ex. A pair of dice is rolled. What is the
probability of rolling
# of possible outcomes
a. A six on the first
c. To win in this situation I must roll 2
AND a five on the second?
numbers (2 blanks)
b. A three on the first
AND a three on the second?
P(rolling an even AND an even) = _____________
c. An even number on the first AND
an even on the second?
d. A 3 on the first and a 3 on the second
OR a 1 on the first and a 1 on the
P(rolling an even AND an even) = 1/4
second.
And or Or
In Probability, the words ‘and’ and ‘or’ are of huge importance.
‘And’ means that
BOTH events occur.
‘Or’ means that ONE
OF the events occur.
# of ways to have success
Ex. A pair of dice is rolled. What is the
# of possible outcomes
probability of rolling
d. To win in this situation I must roll 2
d. A 3 on the first and a 3
on the second OR a 1
on the first and a 1 on
the second.
numbers (2 blanks). I win if I roll {a 3 AND a
3} OR if I roll {a 1 AND a 1}
P(rolling a pair of 3s OR a pair of 1s) =_______
P(rolling a pair of 3s OR a pair of 1s) = 1/18
_____
And or Or
What we have seen is that, FOR A MULTI-STAGED EVENT, ‘and’
means ‘multiply’ and ‘or’ means ‘add’.
Ex 4. A die is rolled and a card is
randomly drawn from a deck. What
is the probability of
a. Rolling a 6 and drawing
a heart?
b. Rolling a 5 and drawing
the 7 of clubs?
c. Rolling a 6 or drawing a
heart?
d. Rolling an odd number
or drawing a queen?
And or Or
What we have seen is that, FOR A MULTI-STAGED EVENT, ‘and’
means ‘multiply’ and ‘or’ means ‘add’.
Ex 4. A die is rolled and a card is
randomly drawn from a deck. What
is the probability of
# of ways to have success
# of possible outcomes
a. P(6 AND Heart) = 1
a. Rolling a 6 and drawing
a heart?
b. Rolling a 5 and drawing
the 7 of clubs?
c. Rolling a 6 or drawing a
heart?
d. Rolling an odd number
or drawing a queen?
13
6 52


1 1

6 4

1
24
And or Or
What we have seen is that, FOR A MULTI-STAGED EVENT, ‘and’
means ‘multiply’ and ‘or’ means ‘add’.
Ex 4. A die is rolled and a card is
randomly drawn from a deck. What
is the probability of
a. Rolling a 6 and drawing
a heart?
b. Rolling a 5 and drawing
the 7 of clubs?
c. Rolling a 6 or drawing a
heart?
d. Rolling an odd number
or drawing a queen?
# of ways to have success
# of possible outcomes
b. P(5 AND 7 of clubs) 

1 1

6 52
1
312
And or Or
What we have seen is that, FOR A MULTI-STAGED EVENT, ‘and’
means ‘multiply’ and ‘or’ means ‘add’.
Ex 4. A die is rolled and a card is
randomly drawn from a deck. What
is the probability of
a. Rolling a 6 and drawing
a heart?
b. Rolling a 5 and drawing
the 7 of clubs?
c. Rolling a 6 or drawing
a heart?
d. Rolling an odd number
or drawing a queen?
# of ways to have success
# of possible outcomes
c. P(6 OR heart) 
HOWEVER, some
of the times that
we rolled a six, we
would have also
drawn a heart. We
cannot count
these successes
twice!
1 13

 P(6 and heart )
6 52

1 13  1 13 
   
6 52  6 52 

1 1 1 1
   
6 4 6 4

3
8
P(A or B)=P(A)+P(B)-P(A and B)
Let’s take a closer look. Consider a party where we dropped
a piece of buttered toast and threw a dart (with our eyes
closed). What is the probability of the toast landed on the
buttered side OR throwing a bull's-eye?
Trial Landed on Bull’s-eye?
So what is P(buttered or bull’s-eye)?
butter?
N
1 N
# of ways to have success
# of possible outcomes
P(buttered or bullseye )  9
9
What a party game! I’m
guaranteed to win!
But wait! I’ve count
some of my wins
twice!
2
Y
3
N
Y
4
N
N
5
N
Y
6
7
Y
8
Y
Y
9
Y
N
Y
N
N
Y
P(A or B)=P(A)+P(B)-P(A and B)
Let’s take a closer look. Consider a party where we dropped
a piece of buttered toast and threw a dart (with our eyes
closed). What is the probability of the toast landed on the
buttered side OR throwing a bull's-eye?
Trial Landed on Bull’s-eye?
So what is P(buttered or bull’s-eye)?
butter?
N
1 N
# of ways to have success
# of possible outcomes
P(buttered or bullseye )  9
9
So I must subtract 2 from
my wins count. This
accounts for the buttered
AND bullseye
P (buttered or bullseye ) 
2
Y
3
N
Y
4
N
N
5
N
Y
6
7
7
9
Y
8
Y
Y
9
Y
N
Y
N
N
Y
And or Or
What we have seen is that, FOR A MULTI-STAGED EVENT, ‘and’
means ‘multiply’ and ‘or’ means ‘add’.
# of ways to have success
Ex 4. A die is rolled and a card is
randomly drawn from a deck. What
is the probability of
a. Rolling a 6 and drawing
a heart?
b. Rolling a 5 and drawing
the 7 of clubs?
c. Rolling a 6 or drawing
a heart?
d. Rolling an odd number
or drawing a queen?
# of possible outcomes
d. P(odd or queen) = P(odd) + P(queen) – P(odd and queen)

3 4 3 4 
   
6 52  6 52 
1 1 1 1 
   
2 13  2 13 
7

13

P(A or B)=P(A)+P(B)-P(A and B)
Let’s take a closer look. Consider an experiment where we
pulled socks from a drawer. 7 socks are blue, 7 are white and
9 are striped.
There are only 19 socks in the drawer, though. How is this possible?
4 of the blue socks are striped!
We can use a Venn diagram
to show this clearly.
blue
3
Since we are pulling only ONCE, we
count the successful events.
P (blue and striped ) 
striped
4
a. P(blue and striped)=?
5
7
4
19
P(A or B)=P(A)+P(B)-P(A and B)
Let’s take a closer look. Consider an experiment where we
pulled socks from a drawer. 7 socks are blue, 7 are white and
9 are striped.
There are only 19 socks in the drawer, though. How is this possible?
4 of the blue socks are striped!
We can use a Venn diagram
to show this clearly.
blue
3
striped
4
5
b. P(blue or striped)=?
Since we are pulling only ONCE, we
count the successful events.
P(blue or striped ) 
P(blue)  P( striped )  P(blue and striped )
7 9 4
P(blue or striped )   
7
19 19 19
12
P(blue or striped ) 
19
Perms and Combos
What is the probability of winning the lotto 6-49?
This type of probability question is one where you’re picking a small group
from a big group (ie. A small group of 6 numbers, from a big group of 49
numbers).
# of ways to have success
# of possible outcomes
So, how many possible outcomes are there?
When I’m drawing blanks, draw blanks
49
x
48
x
47
x
46
This type of calculation can be
simplified using factorials.
x
45
x
44
= 1 x 1010
Factorials
6 factorial is 6x5x4x3x2x1 = 720.
It is written as 6!
10! = 10x9x8x7x6x5x4x3x2x1
10! = 3628800
What is
12!
5!
12! 12 1110  9  8  7  6  5  4  3  2 1

5!
5  4  3  2 1
12! 12 1110  9  8  7  6

 3991680
5!
Factorials
So what is 14 x 13 x 12 x 11 in factorial notation?
Is seems to be 14! But it’s
missing 10!
14!
10!
Factorials are very useful when we’re picking a small group from a big group.
Ex. How many ways are there to randomly select 5 positions out of a group 7 people?
Small group (5) from a big group (7)
7
x
6
x
5
This can be written as 7!
2!
x
4
x
3
To simplify this even further, we
say
7
P5
Perms
When selecting a small group from a big group and the
order selected is important, permutations are used.
Ex2. How many ways can I pick a
president, vice-president from a group of
3.
Group of 3 = A, B, C
Pres
VP
A
B
B
C
n!
n Pr 
(n  r )!
A
C
C
n = # in the big group
r = # in the small group
A
B
3!
6
3 P2 
(3  2)!
Perms
Ex3. a group of 8 books must be arranged on a shelf.
How many possible arrangements are there?
The word ‘arranged’ means that order counts.
I’m picking a ‘small’ group of 8 out of a ‘big’ group of 8 and
order matters.
8!
8!
  40320
8 P8 
(8  8)! 0!
Notice that 0! =1
Combos
When selecting a small group from a big group and the
order selected is not important, combinations are used.
n!
n Cr 
r!(n  r )!
n = # in the big group
r = # in the small group
3!
3  2 1

3
3 C2 
2!(3  2)! (2 1)(1)
Ex. How many ways can 2 people
be picked from a group of 3?
Group of 3 = A, B, C
A
B
B
C
A
C
C
A
B
But, AB = BA so
really there are
only these
options:
AB or CB or AC
Combos
n!
n Cr 
r!(n  r )!
Ex. In a certain poker game, a player is dealt
5 cards. How many different possible hands
are there?
Small group from a big group,
when order doesn’t matter: combo
Big group: 52
Small group: 5
52!
 2598960
52 C5 
5!(52  5)!
4
1
P(royal flush ) 

2598960 649740
So what is the probability of
getting a royal flush
(A,K,Q,J,10 of 1 suit)?
# of ways to have success
# of possible outcomes
There is one royal flush for every
suit so that’s 4 successes.
Perms, Combos and Probability
Ex. A group consists of 7 women and 8 men. A group of 6 must be chosen for a
committee. What is
a. P(exactly 6 women are chosen)?
Small group of 6 from big
group of 15, order doesn’t
matter so it’s a combo.
# of ways to choose 6 women
P(6 women) 
# of ways to choose 6 people
C6
P (6 women) 
15 C6
7
7
1


5005 715
Perms, Combos and Probability
Ex. A group consists of 7 women and 8 men. A group of 6 must be chosen for a
committee. What is
b. P(exactly 4 men are chosen)?
Remember, 6 people are
chosen, so if exactly 4 are
men, 2 must be women.
Small group of 6 from big
group of 15, order doesn’t
matter so it’s a combo.
P(4 men AND 2 women) 
# of ways to choose 4 men AND 2 women
# of ways to choose 6 people
C 4 7 C 2
70  21 1470 42
P(4 men AND 2 women) 



C
5005 5005 143
15 6
8
Perms, Combos and Probability
Ex. A group consists of 7 women and 8 men. A group of 6 must be chosen for a
committee. What is
Remember, ‘at most’ means it could be 1
c. P(at most 2 men are chosen)? man AND 5 women OR 2 men and 4 women
OR no men and 6 women.
P(1 man AND 5 women) 
# of ways to choose 1 man AND 5 women
# of ways to choose 6 people
C  C
P(1 man AND 5 women)  8 1 7 5
15 C6
P(2 men AND 4 women) 
P(2 men AND 4 women) 
P(0 men AND 6 women) 
8
# of ways to choose 2 men AND 4 women
# of ways to choose 6 people
C 2 7 C 4
15 C6
# of ways to choose 0 men AND 6 women
# of ways to choose 6 people
C  C
P(0 men AND 6 women)  8 0 7 6
15 C6
Perms, Combos and Probability
Ex. A group consists of 7 women and 8 men. A group of 6 must be chosen for a
committee. What is
Remember, ‘at least’ means it could be 1
man AND 5 women OR 2 men and 4 women.
c. P(at least 2 men are chosen)?
P(0 m AND 6 w OR 1m AND 5w OR 2m AND 4w)=
8
C0 7 C6 8 C1 7 C5 8 C2 7 C4


15 C6
15 C6
15 C6
P(0 m AND 6 w OR 1m AND 5w OR 2m AND 4w)=
1155
5005
3

13