Transcript Chapter 6: Probability : The Study of Randomness

Chapter 6:
Probability : The Study of Randomness
“We figured the odds as best we could, and
then we rolled the dice.”
US President Jimmy Carter
June 10, 1976
6.1 Randomness
(p. 310-316)

Random phenomenon

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Probability


An outcome that cannot be predicted
Has a regular distribution over many repetitions
Proportion of times that an event occurs in many
repeated events of a random phenomenon
Independent events (trials)

Outcome of an event (trial) does not influence the
outcome of any other event (trial)
Examples:

Independent event

Rolling a single die twice

Result on second roll is
independent of the first

Event that is not
Independent

Picking two cards from a
deck (one at a time with
no replacement)

If you pick a card and do
not replace it and
reshuffle the deck, then
the probability of a red
card on the second pick
is dependent upon what
the first card happened
to be.
6.2 Probability Models
(P. 317-337)
It is often important and necessary to provide a
mathematical description or model for
randomness.

Sample space

The set of all possible outcomes of a random
phenomenon
Example of a Sample Space
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Consider the SUM obtained when two dice are
rolled.
Create a table to display the sums that can be
obtained for this event.
The sample space contains _______ outcomes.
S 

Probability Rules

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If A is an event, then the probability P(A) is a
number between o and 1, inclusive.
P(A does not happen) = 1 – P(A)


The complement of A
Two events are DISJOINT of they have no
outcomes in common.


If A and B are disjoint, then P(A or B) = P(A) + P(B)
Sometimes the phrase MUTUALLY EXCLUSIVE is
used to describe disjoint events
Probability Rules:
An Example

In a roll of two dice


Suppose A = rolling a sum of 7 and B = rolling a
sum of 12, the P(A or B) = 6/36 + 1/36 = 7/36 (A
and B are disjoint).
Suppose C = rolling a sum of 7 and D = rolling an
odd sum, the P(C or D) = P(C) + P(D) – P(C and
D) = 6/36 + 18/36 – 6/36 = 18/36 = ½
(C and D are NOT DISJOINT since 7 is an odd
number)
Probability Rules

Two events are INDEPENDENT if
knowing that one occurs does not
change the probability of the other.

If events A and B are independent,
then P(A and B) = P(A) X P(B)
Examples:

Suppose A = getting a head on a first toss and B =
getting a head on a second toss. A and B are
independent.


P(A and B) = (1/2) X (1/2) = ¼
Suppose C = getting a red card by picking a card from a
randomly shuffled deck of cards and D = getting a red
card by picking a second card from the deck in which the
first card WAS NOT REPLACED. C and D are NOT
independent.


P(C) = 26/52.
If the first card is red, then P(C and D) = (26/52) X (25/51).
 If the first card had been placed back into the deck then

P(C and D) = (26/52) X (26/52) because C and D ARE
independent.
Probability formulas are useful, however, sometimes they
are not needed if sample spaces are small and you use
some common sense.
 Consider a family of two children. There are four
possible boy/girl combinations, ALL EQUALLY
LIKELY.
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P(at least one child is a girl) =
P(two girls) =
P(two girls| one child is a girl) =
P(two girls| oldest child is a girl) =
P(exactly one boy | one child is a girl) =
P(exactly one boy | youngest child is a girl) =
Now suppose that a family unknown to
you has two children.

If one of the children is a boy, what is the
probability that the other child is a boy?
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If the oldest child is a boy, what is the
probability that the other child is a boy?
Many intelligent people think that 50% is the
answer to both of the above. You should be
able to show that this is NOT the case.
More about Probability (P. 341-355)

There are basic laws that govern wise and
efficient use of probability.

FOR ANY TWO EVENTS A and B,
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P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = 0 if A and B are disjoint
P(B|A) = P(A and B)/P(A)
Examples Using the Sum of Two Dice
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P(sum = 7 or sum = 11) =
P(sum = 7 and sum = 11) =
P(sum = 7 or at least one die shows a 5) =
P(faces show same number or sum>9) =
P(sum = 6 | one die show a 4) =
P(sum is even | sum>9) =