Chapter 14: More about chance

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Chapter 14 More about chance
Probabilist computes the probability of complex
events by listing all the possible outcomes of
a chance process Example P(getting a total of
2 spots in one roll of 2 dice) ??
OH, fig 1, p. 239 List out all the possible
ways for two dice to fall 6 x 6 36
possible outcomes in which the 2 dice can
fall only 1 outcome that produces a total of
2 spots
can we use the multiplication rule as per P(2nd
die 1 1st die 1) ? Critical issue Are
these two outcomes dependent/independent?
Independent - so multiply their unconditional
probabilities P(1st die 1) 1/6 P(2nd die
1) 1/6 So P(2nd die 1 1st die 1) 1/6
x 1/6 1/36
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Chapter 14 More about chance

• 17th century Italian game of chance bet on the
  total number of spots rolled with 3 dice believed
  that P(9 total spots) P(10 total spots) but
  they found that 10 spots came up more often
• To answer this we would need a 3 dimensional
  figure, but can use similar reasoning and list
  out all the possible ways for the dice to fall
• Galileo did this die 1 is grey 6
  possible outcomes
• die 2 is white for each of these 6, there are 6
  for the white die thus 6 x 6 possible
  outcomes 36 possible for 2 dice
• die 3 is black for each of these 36, there are 6
  possible for the black die thus a total of 6 x
  6 x 6 216 possible outcomes for 3 dice
• recall chance/probability number of
  outcomes of interest total number of
  possible outcomes
• so we have found what the denominator of this
  equation is next step is the numerator

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Chapter 14 More about chance

• 17th century Italian game of chance bet on the
  total number of spots rolled with 3 dice
• believed that P(9 total spots) P(10 total
  spots), but 10 spots came up more often
• recall P number of outcomes of
  interest 216
• next step is the numerator
• list all the possible ways of getting each
  value (9 or 10) spots with 3 dice

9 spots 1 2 6 1 3 5 1 4 4 2 3 4 2 2
5 3 3 3
ways 6 6 3 6 3 1 25 total ways to get 9
10 spots 1 4 5 1 3 6 2 2 6 2 3 5 2 4
4 3 3 4
ways 6 6 3 6 3 3 27 total ways to get 10
1 2 6 1 2 6 1 6 2 2 1 6 2 6 1 6 1
2 6 2 1 6 ways to get 1 2 6
P(total of 9 in 3 dice) 25 11.6
216
P(total of 10 in 3 dice) 27 12.5
216
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Chapter 14 Exercise Sets 1. Look at the Figure
1 OH. A pair of dice is thrown 1,000 times.
What total should appear most often? What
totals should appear least often?
1 2 3 4 5
2. Two draws will be made at random with
replacement from the box (a) How many total
possible outcomes are there? (b) What is the
chance that the sum of the two draws 6?
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Chapter 14 Exercise Sets 3. In the boxes shown
below, each ticket has two numbers. Find the
chance that when a ticket is drawn at random
from each box, the sum of the two numbers is 4?
1 2 1 3 3 1 3 1

1 2 1 3 1 3 3 2
3 3 3 3
1 2 1 3 1 3 3 1
3 2 3 3
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Chapter 14 Addition Rule The probability that
at least one of two specified events will
happen thus either the FIRST OR the SECOND OR
BOTH
The possibility of both is complicated and can be
ruled out in the special case when two outcomes
are mutually exclusive - for example toss a
fair coin once P(head) .50 P(tail)
.50 both cannot occur simultaneously
- if one occurs, the other cannot Or deal a
single card off the top of a well-shuffled
deck P(card a heart)
13/52 P(card a spade)
13/52 both cannot occur simultaneously
- if one occurs, the other cannot
Addition rule applies when two outcomes are
mutually exclusive simply add their individual
probabilities
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Addition rule for mutual exclusivity add the
individual probabilities
Example deal a single card off the top of a
well-shuffled deck P(the card being either a
heart or a spade) ? P(card a
heart) 13/52 1/4 P(card a spade)
13/52 1/4 P(either suit) 26/52
1/2
What about tossing two dice P(getting at least
one ace) ?? Are
these two mutually exclusive? (does getting one
prevent us from getting the other?) NO
they are not - do not just add the probabilities
here we would have 1/6 1/6 1/3 (12/36)
if we added Look at OH, Fig 1, p.239, the
answer is 11/36 Be careful when counting the
probabilities of events that are not mutually
exclusive - adding double counts
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Chapter 14 Exercise Sets 1. Fifty children
went to a party where cookies and ice cream were
served. 12 kids took cookies, 17 ice
cream. True or false 29 children must have had
cookies OR ice cream?
2. Two cards are dealt off the top of a well
shuffled deck. You have a choice (a) win 1.00
of the first card is an ace or the second card is
an ace (b) win 1.00 if at least one of the 2
cards is an ace Which is better?, are they the
same? Explain
3. Two dice will be rolled. The chance that the
first is a 1 1/6 the chance that the second
is a 1 1/6. True or false. The chance that
the first is an ace or the second is an ace is
1/6 1/6 Explain
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Difference between mutual exclusivity and
independence
A pair of events can be mutually
exclusive OR independent
one occurring does not change the probability of
the other occurring
the occurrence of one prevents the other from
occurring
Probabilities can either be added
OR multiplied
P(at least 1 of 2 events will happen) requires
mutual exclusivity asking P(A or B) ?
P(2 things both happen) P(A and B)
unconditional P for independence conditional P
for dependence
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Difference between mutual exclusivity and
independence
Roll a die 6 times
Question P(first roll 1) or P(last roll 1)
?
Options (i) 1/6 1/6 (ii) 1/6 x 1/6 (iii)
neither
Asking for P(of at least 1 of 2 events
happening) - implies ADDITION Mutually
exclusive? NO so we cannot use the addition rule
so option (i) is out Maybe we can multiply the
two Ps? Are they independent? YES but we are
NOT asking for the P(both will happen)
NOT asking for P(A AND B) so we cannot use the
multiplication rule so (ii) is out Answer is
option (iii)
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Difference between mutual exclusivity and
independence
Roll a die 6 times
Question P(first roll 1) and P(last roll 1)
?
Options (i) 1/6 1/6 (ii) 1/6 x 1/6 (iii)
neither
Asking for P(of both events happening) - so
MULTIPLY Independence? Yes and we are asking
for the P(both will happen) asking
for P(A AND B) the P(B A) so we use the
multiplication rule so (ii) is correct
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Difference between mutual exclusivity and
independence
On a well-shuffled deck of cards
Question P(top cardace of spades) OR P(bottom
card ace of spades) ?
Options (i) 1/52 1/52 (ii) 1/52 x 1/52 (iii)
neither
Asking for P(of at least 1 of 2 events
happening) - so ADD Mutually exclusive? YES so
we can use the addition rule so choose option (i)
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Difference between mutual exclusivity and
independence
On a well-shuffled deck of cards
Question P(top cardace of spades) AND P(bottom
cardace of spades) ?
Options (i) 1/52 1/52 (ii) 1/52 x 1/52 (iii)
neither
Asking for P(of both events happening) - so
MULTIPLY Are they independent? NO, dependent so
we need to multiply the Unconditional
P(1st event) by the conditional P(2nd) 1/52 x
0 therefore choose option (iii)
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Questions asking about the P(at
least one of two things happening) IMPLIES
addition P(this OR that)
addition requires mutual exclusivity
If not, DO NOT ADD, but do not multiply either
not asking about P(both happening)
If mutually exclusive ADD
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Questions asking about the P(two
things both happening) IMPLIES multiplication P(th
is AND that)
multiplication requires independence
If independent P(BA) P(B)
If not independent P(BA) is not P(B)
P(A and B) P(A)P(BA)
P(A and B) P(A)P(B)
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Chapter 14 Exercise Sets 1. A large group of
people are competing for a prize. Each is given
a pack of well shuffled cards, and each deals
two cards off the top of the deck. People win
if the the first is the ace of hearts or the
second is the king of hearts. (a) all those
whose first card was the ace of hearts step
forward what proportion do? (b) those in (a)
return to their places all those whose
second card was the king of hearts step
forward what proportion do? (c) do any people
step froward twice? (d) True or false and
explain the chance of winning 1/52 1/52
2. Two cards are dealt off the top of a well
shuffled deck. (a) Find the chance that the
second card is an ace (b) Find the chance that
the 2nd card is an ace given that the 1st is a
king (b) Find the chance that the 1st is a king
and the 2nd is an ace
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The Paradox of the Chevalier De Mere
Compare the following two bets Roll one die 4
times P(at least one ace turns up) ? Roll a
pair of die 24 times P(at least one double ace
turns up) ? Both events thought to be equally
likely as it seems the question is asking for the
P(at least one of a number of events occurring),
implying addition
Roll one die 4 times P(at least one ace
turns up) ? P(1 turns up
in 1 roll) 1/6, so P(at least one ace turns up)
1/6 1/6 1/6 1/6 2/3 Roll two die 24
times P(at least one double ace turns
up) ? P(1 double ace turns
up in 1 roll) 1/36 so P(at least
one double ace turns up) 1/36 1/36 ...
2/3 BUT experience showed that P(at
least 1 ace in 4 rolls of 1 die) gt P(at least 1
double ace in 24 rolls of two dice)
addition REQUIRES MUTUAL EXCLUSIVITY these
events not mutually exclusive
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The Paradox of the Chevalier DE Mere
Question 1 Roll one die 4 times P(at least one
ace turns up) ?
Note our bet is actually only about two possible
outcomes if one ace turns up in four rolls of 1
die we win if no ace turns up in four rolls of 1
die, we lose it is useful to calculate the P for
the opposite outcome and subtract that from
100 P(losing) P(no ace on roll 1 and no ace on
roll 2 and...etc) 4 independent events So P(no
ace on 1 roll) 5/6 and on 4 rolls 5/6 x 5/6 x
5/6 x 5/6 .482 P(winning) 100 - P(losing)
100 - 48.2 51.8
we cannot multiply the unconditional Ps of the
Pwinning) because that gives us the probability
of winning on every roll 1/6 x 1/6
x 1/6 x 1/6 P(ace on roll 1 AND an ace on roll
2 AND etc) P(all 4 rolls produce an
ace)
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The Paradox of the Chevalier DE Mere
Question 2 Roll pair of dice 24 times P(at
least one double ace turning up) ?
again our bet is actually only about two possible
outcomes if one double ace turns up in 24 rolls
of 2 dice we win if no double ace turns up in 24
rolls of 2 dice, we lose it is useful to
calculate the P for the opposite outcome and
subtract that from 100 P(losing) P(no double
aces on 24 rolls) P(no double ace on roll 1
and no double ace on roll 2 and...etc) again
these are independent events So P(no double ace
on 1 roll of 2 dice) 35/36 and on 24 rolls
(35/36)24 .509 P(winning) 100 - P(losing)
100 - 50.9 49.1
again we cannot multiply the unconditional Ps of
the Pwinning) because that gives us the
probability of winning on every
roll (1/36)24 P(all 24 rolls
produce an ace)