Suppose that Joey arrived late once again this morning and had to park in a noparking
zone.
Can you find the probability that Joey will get a parking ticket?
If so, do it.
If you need additional information to find
the probability, explain what is needed
Problem 12:
Two cards are dealt, one after the other, from a shuffled 52card deck.
Why is it wrong to say that the
probability of getting two red cards is (1/2)(1/2) = 1/4?
What is the correct probability of this event?
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Chapter 6:
Probability and Simulation: The Study of Randomness
Probability Review
Disjoint
or Mutually Exclusive
Events do not have
any outcomes in common.
For example, if A is odd
numbers on a sixsided dice and B is even numbers
on a sixsided dice, then if we roll a die we can not
have both event A and event B occur at the same
time
P(A and B) = 0 (no intersection in Venn).
So
the addition rule for disjoint events
gives us
P(A or B) = P(A) + P(B)
Probability Rules for a Probabilistic Model:
1)
Sum of all P(Events) = 1
2)
All probabilities must be
0 < P(Events) < 1
3)
P(Event) + P(Event’s Compliment) = 1
4)
P(certainty) = 1
and
P(impossibility) = 0
5)
P(unusual event) < 0.05 or 5%
number of ways E can occur
Classical Method of Probability:
P(E) =

total possible ways
observations of F
Empirical Method of Probability:
P(F) = 
total observations
A
B
If events are not mutually exclusive
or are not
disjoint, then they have to have some part in
common. As an example C is numbers < 5 on our
sixsided die and D is numbers > 2.
This common
part would be double counted in the addition rule
above; so the general addition rule
takes this in
account:
P(C or D) = P(C) + P(D) – P(C and D)
C
D
2, 4, 6
1, 3, 5
3, 4
1, 2
5, 6
P(A) = 3/6 = 0.5
and P(B) = 3/6 = 0.5
so
P(A or B) = P(A) + P(B) = 0.5 + 0.5 = 1 (not very common, but A and B are compliments!)
P(C) = 4/6 = 0.67 ,
P(D) = 4/6 = 0.67
and P(C and D) = 2/6 = 0.33
so
P(C or D) = P(C) + P(D)
 P(C and D) = 0.67 + 0.67 – 0.33 = 1
(note:
if we don’t remove the double counting then P(C or D) > 1!)
Computing of AtLeast Probabilities:
P( at least one) = 1 – P(compliment of at least one) = 1 – P(none)
At least one:
x ≥ 1
so the complement is x = 0!
Chapter 6:
Probability and Simulation: The Study of Randomness
Probability Review
Two events are independent
if the probability of them occurring is unaffected by the other
occurring.
Simplest example would be rolling two consecutive sixes on a die.
The
probability of rolling the second six is not dependent on rolling the first six.
The
multiplication rule for independent events
is given by:
P(E and F) = P(E) • P(F) (if E and F are independent!)
P(rolling 2 consecutive sixes) = P(rolling 6) • P(rolling 6) = 1/6 • 1/6 = 1/36 = 0.0278
Two events are dependent
if the probability of them occurring is affected by the other
occurring.
For example, if I drew from a deck of cards an Ace, then probability drawing
another Ace from the deck (without replacing the first card) would be affected by drawing
the first Ace.
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 Fall '12
 SonjaCox
 Probability, AP Statistics, Probability theory, Randomness, The Study of Randomness

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