Final Review – Probability
Definition of Probability
Sample Space is = set of all possible outcomes.
Event Space is = set of all outcome of type E.
Probability of event is ( )
.
Overview of Probability
is a subset of . So always have .
Always have ( ) .
( ) means event NEVER happens.
( ) means event ALWAYS happens.
For 0 < P(E) < 1 , P(E) measures how likely the
event is to happen. Closer to 1 = more likely.
Final Review – Probability
Rules of Probability
Let be the sample space.
Rule 1 : ( ) ( ) ( ) .
Rule 2: P(not E) = 1 – P(E).
Rule 3: Events are MUTUALLY
EXCLUSIVE if they have no common outcome.
(e.g., = exactly 2 Heads, = exactly 3 Heads)
P( ) ( ) ( ).
Rule 4: Events are INDEPENDENT
if they do not affect each other.
(e.g., = first die is 3, = second die is 5)
P( ) ( ) ( )
Final Review – Probability
Odds and Probability
The odds in favor of an event are given as
F to U where F = Favorable, U = Unfavorable.
The odds against an event are U to F.
Formulas
Odds in Favor corresponding probability of success
F to U P =
Odds against correspond to probability of failure
U to F 1-P =
Probability of success corresponding Odds in Favor
P =
A to (B – A)
Probability of failure corresponding Odds Against
P =
A to (B – A)
Know how to do these kinds of problems
(1) General Probability Calculations
(from counting method problems).
(2) Probability when roll 1 or 2 dice.
(3) Probability when flip coins.
(4) Probability when deal cards.
(5) Use Rules of Probability.
(6) Calculate odds from probability.
(7) Calculate probability from odds.
We will not have 3 or more dice in a
probability problem on the final. We will
not mix face up and face down cards in
problems on the final.
Sample Problems – Probability and Odds
Question 1: A password consists of 5
characters where a character is an upper or
lower case letter or a digit. For a random
password, what is the probability that all 5
characters will be the same?
Answer: By the counting methods, there
are 26 + 26 + 10 = 62 possible characters.
The sample space is all 5 character strings.
The event space is all strings with only 1
character. By our counting methods
( )( )( )( )( )
( )( )( )( )( )
The probability that a password has 5
identical characters is:
( )
Sample Problems – Probability and Odds
Question 2: Two dice are rolled. What is
the probability of the following outcome?
(A) Doubles are rolled.
(B) The dice differ by 3.
(C) The total of the dice is not 5.
(D) The total of the dice is 2 or 9.
Answer: For counting see next page.
(A) Doubles: E = {(1,1),(2,2),(3,3),
(4,4),(5,5), (6,6)}. ( )
.
(B) Differ by 3: E = {(1,4), (2,5),
(3,6),(6,3),(5,2),(4,1)}. ( )
.
(C) Total not 5: From 4 ways to total 5,
( )
.
(D) Total 2 or 9: ( )
.
Probability and Counting Methods
Example: When you roll 2 dice the
outcome and corresponding totals are:
The possible outcome when roll two dice
(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)
The total on the two dice are:
2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9
5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12
Sample Problems – Probability and Odds
Question 3: Find probabilities for seeing:
(A) Exactly heads when flip coins.
(B) Exactly heads when flip coins.
(C) 5 or more heads when flip 7 coins.
Answer: These probabilities are:
(A)
Since there are ways for coin
flips to turn out and ways for
exactly of coins to be heads.
(B)
. This is (A) ,
(C) The event is (Exactly 5 heads) OR (Exactly
6 heads) OR (Exactly 7 heads). There are
=29 ways for
this to happen. There are a total of = 128
possible outcome when you flip 7 coins. The
probability is
.
Probability and Counting Methods
Question 4: When you flip 3 coins, what is
the probability of seeing the following:
(A) More heads than tails.
(B) Equal number of heads and tails.
(C) More tails than heads.
(D) At least 1 head or at least 1 tail.
Answer: The sample space for 3 coins flipped is
S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The Event spaces are:
, (empty),
, .
The probabilities of the above events are:
( )
. ( )
(Never happens).
( )
( )
(Always happens).
Probability and Counting Methods
Question 5: Deal 3 cards. Probability for:
(A) 3 Clubs.
(B) 3 Queens.
(C) 3 of a kind.
Answer: The number of outcome in sample space
( )( )( ) . There are
13 clubs so the event space for (A) has
( )( )( )
There are 4 suits, so 4 queens and so
( )( )( ) .
There are 13 types of cards so (13)(24) ways to be
dealt 3 of a kind. ( )( )
( )
. ( )
( )
You get the same answers for instead of
in
all of the above calculations. Stick with one or the
other but do not mix them.
Probability and Counting Methods
Question 6: A has
probability of winning;
B is twice as likely as C; C and D are equally
likely. What are the probabilities that:
(1) A loses?
(2) B wins?
(3) A or C wins?
Answer: The probability A loses is
( )
We know ( ) ( ) ( ) ( )
This is
( ) ( ) ( )
This is ( )
So, ( )
, ( ) ( )
, ( )
Thus, the probability that B or C wins is
( ) ( ) ( )
Probability and Counting Methods
Question 7: In dice a natural is a 7 or an 11.
What are the odds for rolling a natural?
What are the odds against rolling a natural?
Answer: Probability of rolling a natural is
P =
.
A Natural has probability
, means that in 9
trials 2 are naturals, 7 = 9 – 2 are not. So
Odds For a Natural: 2 to 7.
Odds Against Natural: 7 to 2.
You could also calculate that the probability
of NOT rolling a natural is
So the
Odds Against Natural are 7 to 2, where the
2 is given by 2 = 9 – 7.
Probability and Counting Methods
Question 8: The odds of winning for each of
the four teams in a tournament are given by
A has odds 2 to 23. B has odds 8 to 17.
C has odds 21 to 79. D has odds 39 to 61.
Who is most likely to win? Least likely?
Answer: From odds, the probabilities are
Team Odds in Favor Probability Win
A 2 to 23 2/25 (25 = 2 + 23)
B 8 to 17 8/25 (25 = 8 + 17)
C 21 to 79 21/100 (100 = 21 + 79)
D 39 to 61 (100 = 39 + 61)
P(A)
P(B)
P(C)
P(D)
Team D is most likely to win, then B and C.
Team A is least likely to win.
