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Title: Probability ( for genius only :D )
Description: This is intro for the probability theorie... All things about probability ( from low levels to high levels ) for students in schools to colleges and master ... Good luck ( math lover ;) )
Description: This is intro for the probability theorie... All things about probability ( from low levels to high levels ) for students in schools to colleges and master ... Good luck ( math lover ;) )
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Introduction to Probability
Probability Examples c-1
Leif Mejlbro
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Leif Mejlbro
Probability Examples c-1
Introduction to Probability
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...
com
3
Introduction to Probability
Contents
Contents
Introduction
5
1
Some theoretical background
6
2
Set theory
10
3
Sampling with and without replacement
12
4
Playing cards
19
5
Miscellaneous
27
6
Binomial distribution
35
7
Lotto
38
8
Huyghens’ exercise
39
9
Balls in boxes
41
10
Conditional probabilities, Bayes’s formula
42
11
Stochastic independency/dependency
48
12
Probabilities of events by set theory
51
13
The rencontre problem and similar examples
53
14
Strategy in games
57
15
Bertrand’s paradox
59
Index
61
www
...
com
We do not reinvent
the wheel we reinvent
light
...
An environment in which your expertise is in high
demand
...
Implement sustainable ideas in close
cooperation with other specialists and contribute to
influencing our future
...
Light is OSRAM
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This topic is not my favourite,
however, thanks to my former colleague, Ole Jørsboe, I somehow managed to get an idea of what it is
all about
...
On the other hand, it will probably also be closer to the way of thinking which is more common among
many readers, because I also had to start from scratch
...
However, the author
has tried to put them on a minimum, hoping that the reader will meet with sympathy the errors
which do occur in the text
...
com
5
1
...
The topic probability is relying on the concept σ-algebra
...
2) If a set A ∈ F, then also its complementary set lies in F, thus A ∈ F
...
e
...
g
...
e
...
n=1
The sets of F are called events
...
2) P (∅) = 0 and P (Ω) = 1
...
g
...
=
n=1
All these concepts are united in the Probability field, which is a triple (Ω, F, P ), where Ω is a (nonempty) set, F is a σ-algebra of subsets of Ω, and P is a probability measure on (Ω, F)
...
2)
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
...
4)
If A1 ⊆ A2 ⊆ · · · ⊆ An ⊆ · · · and A =
+∞
An ,
then
P (A) = lim P (An )
...
n=1
n→+∞
+∞
5)
If A1 ⊇ A2 ⊇ · · · ⊇ An ⊇ · · · and A =
n=1
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...
Some theoretical beckground
Introduction to Probability
Let (Ω, F, P ) be a probability field, and let A and B ∈ F be events where we assume that P (B) > 0
...
P (B)
In this case, Q, given by
Q(A) := P (A | B),
A ∈ F,
is also a probability measure on (Ω, F)
...
360°
thinking
...
360°
thinking
...
deloitte
...
Discover the truth at www
...
ca/careers
© Deloitte & Touche LLP and affiliated entities
...
com
Discover the truth7at www
...
ca/careers
© Deloitte & Touche LLP and affiliated entities
...
D
1
...
e
...
We expand this by saying that n events Aj , j = 1,
...
, n} have that
⎞
⎛
Aj ⎠ =
P⎝
j∈J
P (Aj )
...
We assume that we have a splitting (Aj )j=1 of Ω into events
Aj ∈ F, which means that the Aj are mutually disjoint and their union is all of Ω, thus
+∞
Aj = Ω,
and
Ai ∩ Aj = ∅,
for every pair of indices (i, j), where i = j
...
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...
Set theory
Introduction to Probability
2
Set theory
Example 2
...
, An be subsets of the sets Ω
...
=
i=1
i=1
These formulæ are called de Morgan’s formulæ
...
If x ∈ ( i=1 Ai ), then x does not belong to any Ai , thus x ∈ Ai for every i, and therefore also
in the intersection, so
n
n
Ai
i=1
Ai
...
On the other hand, if x ∈ i=1 Ai , then x lies in all complements Ai , so x does not belong to
any Ai , and therefore not in the union either, so
n
n
Ai
Ai
i=1
...
2
...
=
i=1
Then by taking the complements,
n
n
Bi =
i=1
Bi
i=1
We see that (2) follows, when we replace Bi by Ai
...
2 Let A and B be two subsets of the set Ω
...
Prove that
AΔB = (A ∪ B) \ (A ∩ B)
...
Prove that
(AΔB)ΔC = AΔ(BΔC)
...
com
9
2
...
The claim is easiest to prove by a Venn diagram
...
If x ∈ (A \ B) ∪ (B \ A), then x either lies in A, and not in B, or in B and not in A
...
1b
...
1c
...
Summing up we get
AΔB = (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)
...
If x ∈ AΔB, then x either lies in A or in B, and not in both of them
...
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...
Set theory
Introduction to Probability
A
B
D
C
Figure 2: Venn diagram of three discs A, B, C
...
(b) If instead x ∈ (AΔB)ΔC and x ∈ C, then x does not belong to AΔB, so either x does not
belong to any A, B, or x belongs to both sets, so we obtain with equality,
{(AΔB)ΔC} ∩ C = {C \ (A ∪ B)} ∪ {A ∪ B ∪ C}
...
By interchanging the letters we get the same right hand side for AΔ(BΔC), hence
(AΔB)ΔC = AΔ(BΔC)
...
com
11
3
...
1 There are 10 different pairs of shoes in a wardrobe
...
Find
the probability of the event that there is at least one pair among them
...
20
4
different
We shall below give two correct and one false solution
...
1) Take the complements, i
...
we apply that
P {at least one pair} = 1 − P {no pair}
...
First choice:
20 possibilities among 20 shoes:
20
,
20
Second choice:
18 possibilities of 19 shoes:
18
,
19
(1 pair not allowed),
Third choice:
16 possibilities of 18 shoes:
16
,
18
(2 pairs not allowed),
Fourth choice:
14 possibilities of 17 shoes:
14
,
17
(3 pairs not allowed)
...
Send us your CV
...
Send us your CV on
www
...
com
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...
Sampling with and without replacement
Introduction to Probability
Summing up,
P {no pair} =
16 · 14
224
20 18 16 14
·
·
·
=
=
,
20 19 18 17
19 · 17
323
hence
P {at least one pair} = 1 −
99
224
=
...
The four shoes stem from 4 pairs, where
10
4
• we can choose 4 pairs in
ways
...
Then
P {no pair} =
10
· 24
4
224
1·2·3·4
10 · 9 · 8 · 7 4
·2 ·
=
...
323
323
2) Direct computation
...
We compute separately the two probabilities on the right hand side
...
The pair can be chosen in
10
1
= 10 ways
...
Then, among the remaining pairs we can choose two in
Within the latter two pairs we choose 1 shoe in
10 ·
P {precisely one pair} =
9
2
20
4
2
1
· 22
= 10 · 4 ·
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...
= 2 ways
...
1 · 2 20 · 19 · 18 · 17
323
3
...
10
2
Two pairs can be chosen in
ways
...
1 · 2 20 · 19 · 18 · 17
323
Summing up it follows by an addition that
P {at least one pair} =
96
3
99
+
=
...
The following is a frequently occurring wrong argument:
10
“P {at least one pair}” =
18
2
20
4
=
102
323
=
6
19
...
Example 3
...
Choose by chance 2r shoes, where
2r < n
...
We have in total 2n shoes, so we have
2n
2r
possibilities
...
(2r)!
1) In this case the 2r shoes must come from 2r pairs
...
Hence
n(2r) · 22r
...
com
14
n
2r
ways
...
Sampling with and without replacement
Introduction to Probability
2) The pair can be chosen in
n
1
= n ways, where we must use two draws
...
= 2 ways, in total 2r − 2 ways
...
(2n)(2r)
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3
...
3 There are 12 parking places in a line on a parking space
...
Find the probability that these 4 places are successive in the line
...
The 4 places can be chosen in
12
4
= 495 ways
...
, (9, 10, 11, 12),
hence
P1 {4 successive places} =
1
9
=
...
, (9, 10, 11, 12), (10, 11, 12, 1), (11, 12, 1, 2), (12, 1, 2, 3)
...
495
165
Example 3
...
1
...
2
...
3
...
1a
...
If 0 is allowed as the first digit, we have all together 105 possibilities
...
All digits can be different in 10 · 9 · 8 · 7 · 6 ways, hence
p1 =
10 · 9 · 8 · 7 · 6
≈ 0
...
105
2
...
The common digit can be chosen in 10 ways
...
5040
...
com
16
3
...
There are only 10 ways, hence
p3 =
10
= 0
...
105
1a
...
Furthermore, if we shall find the probability
of that the 5 digits are different, then we have 9 possibilities for the first place, and 9 − 1 + 1 = 9
possibilities for the second place (because we now can allow 0)
...
3024,
9 · 104
which is the same result as in (1)
...
5 Let n > 3
...
, n, in a sequence
(without replacing the numbers), until they have all been taken
...
2) Find the probability that the numbers 1, 2 and 3 are chosen successively in the given order
...
First notice that the n numbers can be chosen in
n! different orders
...
In this way we
“fix” two places, so we have in reality only n − 1 places to our disposition, hence we have
(n − 1)! possibilities
...
Summing up,
1) P {(1, 2) in the given order} =
1
(n − 1)!
=
...
n!
n(n − 1)
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...
Sampling with and without replacement
Introduction to Probability
Example 3
...
He collects the
2n pieces at random in n pairs
...
When he breaks all matches he gets in total 2n pieces, of which n are long and the remaining n are
short
...
Let us now consider the successes
...
The next piece much be chosen from the n “complementary” pieces, giving n possibilities
...
The fourth piece must be chosen among the remaining n − 1 “complementary” pieces
...
Summing up we get
P {each pair consiste of a short and a long piece} = pn =
2n · (n!)2
=
(2n)!
2n
2n
n
...
10 · 9 · 8 · 7 · 6
63
Example 3
...
Find the probability that the
passengers get off the lift at 7 different storeys
...
The total number of possibilities is 107
...
Hence
P {7 different storeys} =
10 · 9 · 8 · 7 · 6 · 5 · 4
=
107
10
7
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...
06048
...
Playing cards
Introduction to Probability
4
Playing cards
Example 4
...
Find the probability that South has precisely 2 aces
...
We shall deal 4 aces and 35 other cards among the three remaining players
...
3082
...
com
19
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4
...
2 Playing bridge, one of the pairs of partners has in total 9 hearts
...
This example can be solved in many ways
...
First variant
...
1) P {2–2}:
26
ways
...
11
4
c) The 4 hearts are distributed in two talons of 2 in each in
ways
...
11!11!
26!
26 · 25 · 24 · 23
575
2) P {3–1}:
26
ways
...
c) The 4 hearts are distributed in two talons of 3 in one of them and 1 in the other one in
4
4
=
= 4 ways
...
Hence, we have 2
possibilies
...
10!12!
26!
26 · 25 · 24 · 23
575
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...
Playing cards
Introduction to Probability
3) P {4–0}
...
1
Summing up we get
4
0
P {4–0} = 2 ·
22
9
=
26
13
2 · 13 · 12 · 11 · 10
55
11
2 · 22! 13!13!
·
=
=
=
...
♦
P {2–2} + P {3–1} + P {4–0} =
Second variant
...
1) P {2–2}
...
·
2 · 22
1
26 · 25 · 24 · 23
2 · 25 · 23
575
2) P {3–1}
...
1
1·2·3
26 · 25 · 24 · 23
25 · 23
575
3) P {4–0}
...
1·2·3·4
26 · 25 · 24 · 23
575
115
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...
Playing cards
Introduction to Probability
Third variant
...
1
13
=
...
2
...
3
...
4
...
A heart is placed at S with probability
S and N can be chosen in
P {2–2} = 6 ·
4
2
= 6 ways, thus
1 12 13 12
234
·
·
·
=
...
2 25 24 23
575
3) Finally, and also analogously,
P {4–0} = 2
4
4
·
55
11
1 12 11 10
·
·
·
=
=
...
3 Assume that North and South together have 10 trumps
...
East and West have 26 cards together, of which 3 are trumps
...
100
By symmetry, this is also the probability for West having all three trumps
...
100
50
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...
Playing cards
Introduction to Probability
Example 4
...
Find the probability that the four aces
are succeeding each other
...
We see that 4 in a row: (1, 2, 3, 4), (2, 3, 4, 5),
...
Then
P {4 aces} =
49
52
4
=
1
≈ 0
...
5525
Example 4
...
Find the probability in 7 games that at least one of these 7 games have this uniform distribution of
the aces, and find also the probability that precisely one of the 7 games has this uniform distribution
of the aces
...
13! 13! 13! 13!
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...
Playing cards
Introduction to Probability
If we assume that each player has precisely one ace and 12 other cards, this can be done in
4! ·
48!
ways
...
1055
...
4582,
hence
P {at least one of the 7 games has the uniform distribution} = 1 − (1 − p) 7 = 0
...
Finally,
P {precisely 1 of the 7 games has this uniform distribution} = 7 · p(1 − p) 6 = 0
...
Alternatively one may consider 52 places divided into 4 blocks of 13 in each of them
...
Then consider the distribution of the aces between the blocks
...
We choose in each block precisely 1 of the 13 places, so the number of possibilities is
g1 =
13
1
4
= 134 ,
hence,
p1 = P {1 − 1 − 1 − 1} =
134 · 4!
g1
=
= 0
...
m
52 · 51 · 50 · 49
The distribution 2-1-1-0
...
It remains 3 possibilities for
choosing the block, which does not contain any ace
...
5843
...
com
24
4
...
The number of successes is
2
13
2
g3 =
2
13
0
·
4
2
·
= 36 504,
thus
p3 = P {2 − 2 − 0 − 0} =
g3
= 0
...
m
The distribution 3-1-0-0
...
1648
...
The number of successes is
g5 = 4 ·
13
4
·
13
0
4
= 2 860,
hence the probability becomes
p5 = P {4 − 0 − 0 − 0} =
g5
= 0
...
m
Control
...
1055 + 0
...
1348 + 0
...
0106 = 1
...
♦
Example 4
...
Find the smallest
number of games n, for which the probability of North having 4 aces in at least one of the n games is
1
bigger than
...
The successes are characterized by 4 aces and 9 other cards, so there are
4
4
Thus, the probability for North obtaining 4 aces in one game is
p=
48
9
52
13
=
13 · 12 · 11 · 10
11
11
48! 13! 39!
=
=
=
= 0
...
52! 39! 9!
52 · 51 · 50 · 49
49 · 17 · 5
4165
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...
4
...
997359,
4165
4165
so
P {N never obtains 4 aces in n games} = 0
...
This probability is <
1
, when
2
n · ln 0 − 997359 < − ln 2
(< 0),
i
...
when
n>
− ln 2
= 267, 10
...
Remark 4
...
4988,
and
4154
4165
262
= 0
...
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5
...
1 Find the probability of the event that the birthdays of 6 randomly chosen persons are
distributed in precisely 2 months (i
...
such that precisely 10 of the months do not contain any birthday
...
The 6 birthdays can be distributed in 126 ways on the 12 months
...
The 6 birthdays can be distributed in 26 = 64 ways inside the 2 months where, however, 2 ways are
not allowed, because we do not permit that all anniversaries lie only lie in 1 month
...
As a conclusion the probability becomes
11 · 31
341
66 · 62
=
= 5 ≈ 0
...
6
5
12
12
12
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...
Miscellaneous
Introduction to Probability
Alternatively we may apply the following tedious argument:
The probability of the first two persons having their birthdays in different months, while the remaining
four persons have their birthdays in the thus determined months is
12 11
·
·
12 12
4
2
12
=
11 1
·
...
The probability that the first three persons have their birthdays in the same month, the fourth person
in a different month, while the remaining two persons have their birthdays in the thus determined
months is
1·
1
12
2
·
11
·
12
1
12
2
...
12 12
The probability that the first five persons have their birthdays in the same month, while the remaining
person has his birthday in any other month is
1·
1
12
4
·
11
...
com
28
2
+
1
12
3
·
11 2
·
+
12 12
1
12
4
·
341
11
= 5
...
Miscellaneous
Introduction to Probability
Example 5
...
ee
1) Four dices are thrown once
...
2) Now, perform 24 throws with 2 dices
...
2
A French gambler, Chevalier de M´r´, believed that these two probabilities should be equal to , so he
e e
3
lost a lot of money on betting on this
...
1) We get from
P4 {no sixes} =
5
6
4
,
that
P4 {at least one six} = 1 − P4 {no sixes} = 1 −
5
6
4
=1−
671
625
=
≈ 0
...
1296
1296
2) In the same way we get
P24 {no double sixes} =
24
35
36
,
hence
P24 {at least one double sixes} = 1 − P24 {no double sixes} = 1 −
35
36
24
≈ 0
...
It follows immediately that the two probabilities are different
...
1 The result can also be computed directly
...
6
1296
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...
Miscellaneous
Introduction to Probability
Example 5
...
In a town of n + 1 inhabitants one person is telling a rumour to another one, who tells the story to
another one etc
...
1) Find the probability of that the romour is told r times without returning to the person who started
gossiping
...
It is here allowed that the rumour can also be told to the person who has just told it to himself, cf
...
4
...
The following persons have n − 1 choices of n possibilities, thus the probability is
1−
pr =
1
n
r−1
for r ≥ 2,
and we see that the formula is trivial for r = 1, hence it is valid for all r ∈ N
...
Person
number j has n + 1 − j possible successful choices
...
n
n
n
n
(n − r)! n
Example 5
...
3, by assuming that a rumour-monger never
chooses to tell the rumour to the same person who just told it to himself
...
3 the probability is 1 for r = 1
...
Thus we get the probability
1−
pr =
1
n−1
1
, because
n−1
r−1
for r > 1,
and hence for all r ∈ N
...
2) The first person can choose among n persons
...
Thus
q1 = 1
and
q2 = 1
...
n−1 n−1
n−1
(n − r)! (n − 1)r−2
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...
Miscellaneous
Introduction to Probability
Example 5
...
Find the probability that all
pupils have different birthdays (we assume that all 365 days have the same probability), and find
approximative values of pr for reasonable small values of r
...
365 365 365
365
Approximative values for r = 1,
...
000,
p2 = 0
...
992,
p4 = 0
...
973,
p6 = 0
...
944,
p8 = 0
...
905,
p10 = 0
...
859,
= 0
...
806,
= 0
...
747,
= 0
...
685,
= 0
...
621,
= 0
...
556,
= 0
...
493,
= 0
...
431,
= 0
...
373,
= 0
...
319,
= 0
...
Remark 5
...
then we have
the probability 1 − 0
...
507 that at least two pupils have the same birthday
...
ligsuniversity
...
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5
...
6 Assume that there are 100 lots in a lottery and only one prize
...
1) 1st variant
...
0200
...
p
= P {prize on the 1st lot} + P {loss on the 1st lot and prize on the 2nd lot}
1
99 1
1
1
2
=
+
·
=
+
=
= 0
...
100 100 99
100 100
100
2) 1st variant
...
9801 = 0
...
2nd variant
...
0199
...
0001) then just 1 prize, the average
can only remain the same if we also lower the probability in (2), hence the probability of (2) is smaller
than the probability of (1)
...
com
32
5
...
7 There are two roads from X-town to Y-town, and two roads from Y-town to Z-town
...
One day there is
a blizzard over the area, and we assume that each of the four roads are closed with the probability p
...
Find the probability that there is an open connection from X-town to Z-town
...
The probability of the event that both roads from X to Y are closed is p2
...
Analogously for the roads from Y to Z
...
com
33
2
...
Miscellaneous
Introduction to Probability
and analogously
P {A gets from Y to Z without problems} = 1 − p
...
2
Obviously, when 0 < p < 1, then (1 − p)2 < 1 − p2
...
8 Given a roulette, where the outcome of a single game is either red of probability p, or
black of probability q = 1 − p
...
We denote by X the stochastic variable which indicates the number of games in the first unchanged
sequence of games
...
1) Find for each n ∈ N the probability of X = n
...
3) Find the minimum of the mean as a function of p
...
2) The mean value is computed in the usual way:
∞
E{X}
∞
n P {X = n} = pq
=
n=1
= pq
3) If we put x =
n pn−1 + n q n−1
n=1
1
1
+ 2
2
q
p
p q
= +
...
The minimum is obtained for x =
p
1
= 1, corresponding to p =
...
If e
...
The result is reasonable because we for p =
2
1
p > , then we shall get more reds in a row
...
com
34
6
...
1 A coin is thrown 2n times, where we assume that each of the 22n possibilities have the
same probability 2−2n
...
2) Find the probability that we obtain heads precisely 2 times
...
1) We apply the binomial distribution with p = q =
P {X = n} =
2n
n
·
1
, so
2
1
...
2) Analogously,
P {X = 2} =
2n
2
·
2n(2n − 1) 1
n(2n − 1)
1
· 2n =
=
...
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...
Binomial distribution
Introduction to Probability
3) Considering the complementary event we get
2n
0
P {X ≥ 2} = 1 − P {X = 0} − P {X = 1} = 1 −
= 1 − (1 + 2n)
·
1
−
22n
2n
1
·
1
22n
1
2n + 1
=1−
...
2 Some airplane companies allow for that some of their passengers who have booked a
ticket do not show up at take off
...
1
for not showing up and that
We assume in the following that each passenger has the probability of
10
this happens independently of the other passengers
...
Company B always sells 20 tickets to its larger airplane of 18 seats
...
Find the probability that company B at take off must reject 1 or 2 passengers
...
We apply the binomial distribution, so
10
k
P {k passengers appear} =
k
9
10
·
1
10
10−k
,
k = 0, 1,
...
Hence
10
10
9
10
= 0
...
Company B
...
, 20,
hence k successes, 20 − k failures and the probability of success,
9
18
=
...
3917
...
If one company sells 100 tickets with only seats for 90, the probability of rejection is
0
...
If it sells 400 tickets with seats for 360, the probability of rejection is
0
...
We see that the larger the airplane, the higher the probability of rejection
...
2
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...
Binomial distribution
Introduction to Probability
Example 6
...
3
Find the probability for A obtaining 13 successes, and the probability for A obtaining precisely 12
successes
...
Find the probability that B obtains 13 successes, and the probability that B gets precisely successes
...
3
Find the probability that C obtains 13 successes, and the probability that C gets precisely 12 successes
...
1) The probability of success is p =
PA {13 successes} =
PA {12 successes} =
1
= 0
...
000 016 3
...
000 244,
213
1
13
13
· 13 = 13 = 0
...
12
2
2
3) The probability of success is p =
PA {13 successes} =
1
, so
3
2
3
2
, so
3
13
13
12
= 0
...
com
37
2
3
13
= 0
...
7
...
1 In lotto, 7 numbers are drawn among 1, 2,
...
A young hopeful, who wants to
travel around the world, chooses 7 numbers on a lotto coupon
...
There are in total
36
7
= 8 347 680
possible different lotto coupons
...
000 000 12
...
Then we compute the number of possibilities of 6 winning numbers:
1) The six winning numbers can be chosen in
7
6
= 7 ways
...
This can be done
29
in
= 29 ways
...
000 02432
...
, 7,
thus
p0 = 0
...
0153,
p1 = 0
...
0010,
p2 = 0
...
000 024 32,
p3 = 0
...
000 000 12
...
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...
Huyghens’ exercise
Introduction to Probability
8
Huyghens’ exercise
Example 8
...
The game ends, when either A throws
in total 6 (in which case A wins), or B throws in total 7 (in which case B wins B)
...
This exercise occurs in one of the very first books on the calculus of probability, written in 1656 by
Huyghens
...
The table also shows that 5 of
these give the sum 6, and that 6 of them give the sum 7
...
6
P {A wins in a throw} = pA =
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...
Huyghens’ exercise
Introduction to Probability
The gamblers now throw by turns with A first
...
Summing up we get
+∞
P {A wins} =
+∞
P {A wins in throw number 2n + 1} =
n=0
=
1
n
(qA qB ) pA =
n=0
5
36
− 31
36
·
5
6
=
pA
1 − q A qB
30
30
=
,
216 − 155
61
and
+∞
P {B wins}
+∞
P {B wins in throw number 2n + 2} =
=
n=0
=
31 1
36 · 6
1 − 31 ·
36
n
(qA qB ) qA pB =
n=0
5
6
=
31
31
=
,
216 − 155
61
As a check we see that the sum of the probabilities is
30 31
+
= 1
...
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...
Balls in boxes
Introduction to Probability
9
Balls in boxes
Example 9
...
To the time 1 we draw at random 1 ball from each of the two boxes and interchange them
...
Let pn denote the probability that the black ball is in A after n such interchangings
...
4 2
2) Prove that pn =
2n + 1
, n ∈ N
...
1) When n = 2, 3,
...
=
4
4
4 2
2) If n = 1, then
p1 =
3
21 + 1
number of whites in A
= =
...
n
3
4 2
4
2·2
2
23
Hence, the formula is true for n = 1 and for n = 2
...
Then
pn+1 =
1 1
2n
2n + 1
2n + 2n + 1
2n+1 + 1
+ pn = n+2 + n+2 =
=
,
n+2
4 2
2
2
2
2n+2
and the formula is also true for n + 1, and the claim follows by induction
...
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...
Comditional probabilities, Bayes’s formula
Introduction to Probability
10
Conditional probabilities, Bayes’s formula
Example 10
...
Find the
probability that all 3 children are boys
...
There are in total 23 = 8 possibilities
...
Since the 7 events have the same probability, the searched
1
probability is
...
1 One frequently here sees Bayes’s formula applied, in which case one obtains the wrong
1
result
...
g
...
♦
4
Example 10
...
Find the conditional
probability that the first six occurs in the 2nd throw, given that the first six occurs in a throw of even
number
...
If the first six occurs in throw number 2n, then the first 2n − 1 throws cannot have given a six
...
6
In particular,
P {first six in throw number 2} =
5
5 1
· =
...
com
42
11
...
Comditional probabilities, Bayes’s formula
Introduction to Probability
Example 10
...
We draw one card at random from the talon
...
6
2
2) There are 5 aces, 3 of which are red cards, so the probability is
3
...
com
axa_ad_grad_prog_170x115
...
Comditional probabilities, Bayes’s formula
Introduction to Probability
Example 10
...
Box A contains 2 red and 3
black balls, box B contains 1 red and 4 black balls, and box C contains 3 red balls and 1 black ball
...
Assume that the
drawn ball is red
...
Let
denote a red ball and ∗ a black one
...
4
Then by Bayes’s formula,
P {A | red} =
=
P (A) · P {red | A}
P (A) · P {red | A} + P (B) · P {red | B} + P (C) · P {red | C}
1 2
2
2
40
8
3 · 5
5
...
5 Two boxes A1 and A2 contain w1 white and b1 black balls, and w2 white and b2 black
balls, resp
...
Find the probability that this ball is white
...
Since we choose 1 ball from each box, we get
P (Ai ) =
1
,
2
i = 1, 2
...
2 w1 + b1
2 w2 + b2
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...
Comditional probabilities, Bayes’s formula
Introduction to Probability
Example 10
...
One
1
to a 1, and that a sent 1 is changed with the
expect that a sent 0 is changed with the probability
5
2
1
probability to a 0
...
6
3
1) Assuming that we receive a 0, what is the probability that a 0 was sent?
2) Assuming that we receive a 1, what is the probability that a 1 was sent?
We first write down Bayes’s formula,
P (Ak | A) =
P (Ak ) · P (A | Ak )
,
i P (Ai ) P (A | Ai )
i
...
Ak and A “are interchanged”
...
When this is put into Bayes’s formula, we get
P {0 sent
...
| 0 sent} + P {1 sent}P {0 received | 1 sent}
2
3
4
5
2
·
3
·
4
5
+
1
3
·
1
6
=
8
15
8
15
1
+
18
=
48
90
53
90
=
48
53
(≈ 91%)
...
Then by insertion,
P {1 sent | 1 received} =
=
P {1 sent} · P {1 received | 1 sent}
P {1 sent}P {1 received | 1 sent} + P {0 sent}P {1 received | 0 sent}
1
·
3
1
3
5
6
·
+
5
6
2
3
·
1
5
=
5
18
5
18
2
+
15
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...
10
...
7 A factory buys 1000 light bulbs of type A, and 500 bulbs of type B, which are somewhat
more expensive
...
6 of that it lasts longer than 2 months
...
9 that it lasts longer than 2 months
...
A bulb is chosen at random from the 1500 bulbs
...
If a bulb lasts for more than 2 months, what is the probability that it is of type A?
It follows directly that
P {the bulb lasts in more than 2 months}
500
2
1
1000
· 0
...
9 = · 0
...
9 = 0
...
57)
...
6
2·6
4
=
=
0
...
8 Given n slips of paper, each with one of the numbers 1, 2,
...
The n slips are
randomly drawn one by one
...
, n − 1
...
e
...
Then
P { ak = n | ak > a1 , ak > a2 ,
...
, ak > ak−1 }
=
...
, ak > ak−1 }
P {ak > a1 , ak > a2 ,
...
, in the k-th draw, we must have
P {ak > a1 , ak > a2 ,
...
k
Since
P {ak = n} =
1
,
n
we finally get that
P { ak = n | ak > a1 , ak > a2 ,
...
1/k
n
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...
Comditional probabilities, Bayes’s formula
Introduction to Probability
1
Example 10
...
We choose from a given collection of 1000 bulbs a random test consisting of 100 bulbs
...
If we know that there are no defective bulbs in this random test, what is then the probability that there
are no defective bulbs among the 1000 bulbs?
Let Ak denote the event that there are k defective bulbs among the 1000, where k = 0, 1, 2, 3, 4, 5
...
6
Let A be the event that there are no defective bulbs among the 100 bulbs
...
k=0
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...
Comditional probabilities, Bayes’s formula
Introduction to Probability
By a computation,
P (A | A0 ) = 1,
901 900
999 998
·
···
·
=
P (A | A1 ) =
1000 999
902 901
900 899
998 997
·
···
·
=
P (A | A2 ) =
1000 999
902 901
898
· P (A | A2 ) = 0
...
6557,
P (A | A4 ) =
997
896
· P (A | A4 ) = 0
...
9,
·
899
899
=
· P (A | A1 ) = 0
...
7807
...
2135
...
7807
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...
Stochastic independency / dependency
Introduction to Probability
11
Stochastic independency/dependency
Example 11
...
Introduce for i < j the events
Aij by
Aij = {same numbers in the i-th and the j-th throw}
...
2) Prove that the events Aij , where 1 ≤ i < j ≤ n, are pairwise independent, though not independent
...
, 6}, we have the probability of obtaining
6
the same result m in the j-th throw, so
P (Aij ) =
1
...
Clearly, if all four numbers are mutually
different, then clearly
P (Aij ∩ Ak ) = P (Aij ) · P (Ak )
...
j, is equal to one of the numbers k, , the remaining two indices are
still different, so given the number of the common index the probability is
P (Aij ∩ Ak ) =
1 1
· = P (Aij ) · P (Ak )
...
Since n ≥ 3, it suffices to consider the event A12 ∩ A13 ∩ A23
...
Then the event A23 is automatically
6
fulfilled, so
P (A12 ∩ A13 ∩ A23 ) = P (A12 ∩ A13 ) = P (A12 ) · P (A13 ) = P (A12 ) · P (A13 ) · P (A23 ) ,
1
because P (A23 ) = = 1
...
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...
Stochastic independency / dependency
Introduction to Probability
Example 11
...
Each of the 2n possible events are given the
probability 2−n
...
What can be said about the
independency of A and B?
It follows that
P (A) = 1 − P {all tails} − P {all heads} = 1 −
1
1
2n−1 − 1
1
− n = 1 − n−1 =
,
2n
2
2
2n−1
and
P (B) = P {one tail} + P {all heads} =
1
n+1
n
+ n =
...
2n
This expression is equal to
P (A) · P (B) =
2n−1 − 1 n + 1
· n ,
2n − 1
2
if and only if
n + 1 = 2n−1 ,
which happens if (and only if) n = 3
...
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dissertation?
Get in-depth feedback & advice from experts in your
topic area
...
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...
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...
Probabilities of events by set theory
Introduction to Probability
12
Probabilities of events by set theory
Example 12
...
, An denote n events
...
For n = 1 we get the trivial identity P (A1 ) = P (A1 )
...
It follows from
P (A1 ∪ A2 ) = P (A1 \ A2 ) + P (A2 \ A1 ) + P (A1 ∩ A2 )
= {P (A1 \ A2 ) + P (A1 ∩ A2 )} + {P (A2 \ A1 ) + P (A1 ∩ A2 )} − P (A1 ∩ A2 )
= P (A1 ) + P (A2 ) − P (A1 ∩ A2 ) ,
that (2) holds
...
(It follows from the above that it is true for
n = 1 and for n = 2)
...
Then it follows from (2) that
n+1
P
n
Ai
= P
i=1
Ai ∪ An+1
i=1
n
= P
n
Ai
+ P (An+1 ) − P
i=1
P (Ai ) −
=
{Ai ∩ An+1 }
i=1
i≤n
P (Ai1 ∩ Ai2 ) +
i1
P (Ai1 ∩ Ai2 ∩ Ai3 ) + · · ·
i1
= (−1)n−1 P (A1 ∩ A2 ∩ · · · ∩ An ) + P (An+1 )
−
P (Ai ∩ An+1 ) +
i≤n
P (Ai1 ∩ Ai2 ∩ An+1 )
i1
−
P (Ai1 ∩ Ai2 ∩ Ai3 ∩ An+1 )
i1
+ · · · + (−1)n P (A1 ∩ A2 ∩ · · · ∩ An ∩ An+1 )
P (Ai ) −
=
i
P (Ai ∩ Ai2 ) +
i1
P (Ai1 ∩ Ai2 ∩ Ai3 ) + · · ·
i1
+(−1)n P (A1 ∩ A2 ∩ · · · ∩ An+1 ) ,
which is the wanted formula with n replaced by n + 1
...
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...
Probabilities of events by set theory
Introduction to Probability
Example 12
...
, Bn denote n events
...
Hint: The example can either be shown by induction, or one may use that
Bi =
Bi
i∈J
i∈J
for any index set, and then apply the result of Example 12
...
The proof by induction follows the same pattern as the proof of Example 12
...
e
...
1 is known, Then
n
= P
Bi
P
=1−P
Bi
i=1
i∈J
Bi −
P
= 1−
Bi
i∈J
i
P
Bi1 ∩ Bi2
i1
Bi1 ∩ Bi2 ∩ Bi3 + · · · + (−1)n−1 P
P
+
B1 ∩ B2 ∩ · · · ∩ Bn
i1
= 1−
Bi −
P
i
P
(Bi1 ∪ Bi2 )
i1
(Bi1 ∪ Bi2 ∪ Bi3 ) + · · · + (−1)n−1 P
P
+
(B1 ∪ B2 ∪ · · · ∪ Bn )
i1
{1 − P (Bi )} −
= 1−
i
{1 − P (Bi1 ∪ Bi2 )}
i1
{1 − P (Bi1 ∪ Bi2 ∪ Bi3 )} + · · · + (−1)n−1 (1 − P (B1 ∪ B2 ∪ · · · ∪ Bn ))
+
i1
P (Bi ) −
=
i
P (Bi1 ∪ Bi2 )
i1
P (Bi1 ∪ Bi2 ∪ Bi3 ) + · · · + (−1)n−1 P (B1 ∪ B2 ∪ · · · ∪ Bn )
+
i1
1−
=
n
1
+
n
2
n
3
−
+ · · · + 1
...
com
52
1n−j (−1)j = (1 − 1)n = 0
...
The rencontre problem and similar examples
Introduction to Probability
13
The rencontre problem and similar examples
Example 13
...
In an airplane with n seats, n passengers – each provided with a numbered ticket – take at random
their seats in the airplane
...
Hint: Let Ak denote the event that the k-th passenger sits on the right seat
...
1
...
Another formulation is the hat problem: A sleepy
cloakroom attendant distributes n hats between n gentlemen without bothering with whether the hats
are in fact given to their owners; here pn is the probability that no one gets his own hat
...
According to Example 12
...
−
i1
Here, P (Ai ) =
1
, and
n
P (Ai ∩ Aj ) = P (Ai ) · P (Aj | Ai ) =
1
1
·
n n−1
for i < j,
and in general,
P (Ai1 ∩ Ai2 ∩ · · · ∩ Aik ) =
It follows that
i
for i1 < i2 < · · · < ik
...
com
53
13
...
We get by insertion,
pn = 1 −
1
1
1
(−1)n
1
+
−
+
+ ··· +
→ e−1 ≈ 0
...
For n = 1 it is obvious that the probability is p1 = 0
...
375,
8
p5 =
11
≈ 0
...
30
Since pn is a section of the alternating series
∞
n=0
(−1)n
n!
where the absolute value of the n-th term, 1/n!, is obtained by removing the alternating factor (−1) n ,
and which decreases towards zero, the error of the truncated series is always smaller than the first
rejected term, i
...
pn =
1
+ rn ,
e
where |rn | ≤
1
...
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The rencontre problem and similar examples
Introduction to Probability
Example 13
...
The n + m slips are drawn one by one without return
...
Indicate in
particular the probabilities in the case of m = 4, n = 5 and m = 4, n = 8
...
We can choose n among these (the
position of the 1s) in
m+n
n
m+n
m
=
m˚
ader
...
, (0 1), (0)
...
The remaining
n − (m − 1) = n − m + 1,
must be distributed among n + 1 positions
...
This is
equivalent to the placement of the m groups with a 0 on n + 1 positions, which can be done in
m+n
m
ways
...
We get for m = 4 and n = 5, and for m = 4 and n = 8,
6
4
9
4
=
5
,
42
9
4
12
4
=
14
...
3 In lotto, 7 winning numbers and 4 additional numbers are drawn among the numbers
1, 2, 3,
...
It happens quite often that there amongst the 7 winning numbers are two or more
neighbouring numbers, e
...
both 6 and 7 are winning numbers
...
2) Find the probability that there are no neighbouring numbers among the 7 winning numbers
...
Hint: The example is related to the previous example
...
Then we can draw n numbers among 36 in
36
n
ways
...
com
55
13
...
This
gives n constraints, and we get 36 − n possibilities for the n places
...
(b) If 36 is chosen, then we are left with 35 places and n − 1 constraints, hence 35 − (n − 1) = 36 − n
places for n − 1 pairs
...
Summarizing we get
36 − n
n
+
36 − n
n−1
37 − n
n
=
successes,
so the probability is
37 − n
n
36
n
...
694
...
244
...
0129
...
com
56
14
...
1 Andrew has bought 3 unusual dices
...
All 3 dices are considered as true, i
...
be throwing any of the three dices each surface
1
has the probability for showing up
...
Peter is allowed first to choose among the
3 dices, and afterwards Andrew can choose between the two remaining dices
...
Who has won most prizes?
The example reminds very much of the well-known game of “stone–paper–scissors”, so a qualified
guess is that Andrew will win most prizes, because he knows Peters strategy
...
We get by a computation that
P {A > B} = (P {A = 14}+P {A = 13}+P {A = 12}+P {A = 11}+P {A = 6}) ×
×(P {Y = 4} + P {Y = 3} + P {Y = 2})
1
5
1
,
= 5· ·3· =
6
6
12
P {B > C} = (P {B = 17} + P {B = 16} + P {B = 15}) ×
×(P {C = 10}+P {C = 9}+P {C = 8}+P {C = 7}+P {C = 5})
1
5
1
,
= 3· ·5· =
6
6
12
P {C > A} = P {C = 18} · (P {A = 14} + P {A = 13} + P {A = 12} +
+P {A = 11} + P {A = 6} + P {A = 1})
+(P {C = 10} + P {C = 9} + P {C = 8} + P {C = 7}) ×
×(P {A = 6} + P {A = 1})
+P {C = 5} · P {A = 1}
1
1 1 1 1
6+8+1
15
5
1
·6· +4· ·2 + · =
=
=
...
e
...
e
...
e
...
12
Hence, Andrew’s strategy must be:
1) If Peter chooses A, then Andrew chooses B
...
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...
Strategy in games
Introduction to Probability
3) If Peter chooses C, then Andrew chooses A
...
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...
Bertrand’s paradox
Introduction to Probability
15
Bertrand’s paradox
Example 15
...
What is the probability p that the cord
is longer than the side length of an inscribed equilateral triangle in the circle?
B
C
A
Solution 1
...
If it lies on the smallest of the arcs BC, then the cord is longer then the side length of the
1
equilateral triangle, hence p =
...
The random cord is perpendicular so some diameter AD
...
2
Solution 3
...
Then the
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...
Bertrand’s paradox
Introduction to Probability
cord is longer than the side length of the equilateral triangle, if the midpoint lies inside the inscribed
1
r
circle of radius , hence p =
...
The example is known as Bertrand’s paradox
...
In the second solution we use the length on AD, divided by the length of AD, as our probability
measure
...
These three probability measures are clearly mutually different, explaining why we get different results
...
com
60
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Index
Introduction to Probability
Index
Bayes’s formula, 6
conditional probability, 5
draw, 10
event, 4
independent events, 6
law of total probability, 6
de Morgan’s formulæ, 7
multiplication theorem of probability, 5
probability field, 4
probability measure, 4
splitting, 6
symmetric set difference, 7
σ-algebra, 4
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Title: Probability ( for genius only :D )
Description: This is intro for the probability theorie... All things about probability ( from low levels to high levels ) for students in schools to colleges and master ... Good luck ( math lover ;) )
Description: This is intro for the probability theorie... All things about probability ( from low levels to high levels ) for students in schools to colleges and master ... Good luck ( math lover ;) )