5. Probability
• Denoted by P(Event)
outcomestotal
outcomesfavorable
)( =EP
This method for calculating probabilities is only appropriate
when the outcomes of the sample space are equally likely.
6. Experimental Probability
• The relative frequency at which a
chance experiment occurs
–Flip a fair coin 30 times & get 17
heads
30
17
7. Law of Large Numbers
• As the number of repetitions of a
chance experiment increase, the
difference between the relative
frequency of occurrence for an
event and the true probability
approaches zero.
8. Basic Rules of Probability
Rule 1. Legitimate Values
For any event E,
0 < P(E) < 1
Rule 2. Sample space
If S is the sample space,
P(S) = 1
10. Rule 4. Addition
If two events E & F are disjoint,
P(E or F) = P(E) + P(F)
(General) If two events E & F are
not disjoint,
P(E or F) = P(E) + P(F) – P(E & F)
11. Ex 1) A large auto center sells cars made by
many different manufacturers. Three of these
are Honda, Nissan, and Toyota. (Note: these
are not simple events since there are many
types of each brand.) Suppose that P(H) = .
25, P(N) = .18, P(T) = .14.
Are these disjoint events?
P(H or N or T) =
P(not (H or N or T) =
yes
.25 + .18+ .14 = .57
1 - .57 = .43
12. Ex. 2) Musical styles other than rock
and pop are becoming more popular. A
survey of college students finds that the
probability they like country music is .
40. The probability that they liked jazz
is .30 and that they liked both is .10.
What is the probability that they like
country or jazz?
P(C or J) = .4 + .3 -.1 = .6
13. Rule 5. Multiplication
If two events A & B are
independent,
General rule:
P(B)P(A)B)&P(A ⋅=
A)|P(BP(A)B)&P(A ⋅=
14. Ex. 3) A certain brand of light bulbs are
defective five percent of the time. You
randomly pick a package of two such
bulbs off the shelf of a store. What is the
probability that both bulbs are defective?
Can you assume they are independent?
0025.05.05.D)&P(D =⋅=
15. Ex 4)
If P(A) = 0.45, P(B) = 0.35, and A &
B are independent, find P(A or B).
Is A & B disjoint?
If A & B are disjoint, are they
independent?
Disjoint events do not happen at the same
time.
So, if A occurs, can B occur?
Disjoint events are
dependent!
NO, independent events cannot be disjoint
P(A or B) = P(A) + P(B) – P(A & B)
How can you
find the
probability of
A & B?
P(A or B) = .45 + .35 - .45(.35) =
0.6425
If
independent,
multiply
16. Ex 5) Suppose I will pick two cards from a standard
deck without replacement. What is the probability that
I select two spades?
Are the cards independent? NO
P(A & B) = P(A) · P(B|A)
Read “probability of B
given that A occurs”
P(Spade & Spade) = 1/4 · 12/51 = 1/17
The probability of getting a spade given
that a spade has already been drawn.
17. Rule 6. At least one
The probability that at least one
outcome happens is 1 minus the
probability that no outcomes
happen.
P(at least 1) = 1 – P(none)
18. Ex. 6) A certain brand of light bulbs are
defective five percent of the time. You
randomly pick a package of two such
bulbs off the shelf of a store.
What is the probability that at least one
bulb is defective?
P(at least one) = P(D & DC
) or P(DC
& D) or P(D & D)
or
1 - P(DC
& DC
)
.0975
19. Ex 7) For a sales promotion the
manufacturer places winning symbols
under the caps of 10% of all Dr. Pepper
bottles. You buy a six-pack. What is
the probability that you win something?
P(at least one winning symbol) =
1 – P(no winning symbols)
1 - .96
= .4686
20. Rule 7: Conditional Probability
• A probability that takes into
account a given condition
P(A)
B)P(A
A)|P(B
∩
=
given
and
A)|P(B =
21. Ex 6) In a recent study it was
found that the probability that a
randomly selected student is a
girl is .51 and is a girl and plays
sports is .10. If the student is
female, what is the probability
that she plays sports?
1961.
51.
1.
P(F)
F)P(S
F)|P(S ==
∩
=
22. Ex 7) The probability that a
randomly selected student plays
sports if they are male is .31. What
is the probability that the student is
male and plays sports if the
probability that they are male
is .49?
1519.
49.
31.
P(M)
M)P(S
M)|P(S
=
=
∩
=
x
x
23. • The number of possible permutations is the
number of different orders in which
particular events occur. The number of
possible permutations are
Permutations
24. Np=
n!
(n−r)!
where r is the number of events in the series, n is the number
of possible events, and n! denotes the factorial of n = the
product of all the positive integers from 1 to n.
26. Permutations
In how many ways can 4 CD’s (out of a
collection of 8 CD’s) be arranged on a
shelf?
Np=
n!
(n−r)!
n = 8
r = 4
Np=
8!
(8−4)!
=
8!
4!
=1,680
27. When the order in which the events occurred is of no interest, we
are dealing with combinations. The number of possible
combinations is
When the order in which the events occurred is of no interest, we
are dealing with combinations. The number of possible
combinations is
CombinationsCombinations
Nc=
n
r
=
n!
r!(n−r)!
28. where r is the number of events
in the series, n is the number of
possible events, and n! denotes
the factorial of n = the product
of all the positive integers from
1 to n.
29. How many groups of 4 CDs are there in a collection
of 8 CDs?
How many groups of 4 CDs are there in a collection
of 8 CDs?
CombinationsCombinations
Nc=
n
r
=
n!
r!(n−r)!
n = 8
r = 4
Nc=
8
4
=
8!
4!(8−4)!
=
8!
4!4!
=70
