1 independence supplementary notes prepared by raymond wong presented by raymond wong

1

Independence

Supplementary Notes

Prepared by Raymond WongPresented by Raymond Wong

2

e.g.1 (Page 3) The sample space of throwing two fair dice is

Dice 1

Dice 2

1 1

1 2

1 3

1 4

1 5

1 6

2 1

2 2

2 3

2 4

2 5

2 6

Dice 1

Dice 2

3 1

3 2

3 3

3 4

3 5

3 6

4 1

4 2

4 3

4 4

4 5

4 6

Dice 1

Dice 2

5 1

5 2

5 3

5 4

5 5

5 6

6 1

6 2

6 3

6 4

6 5

6 6

3

e.g.2 () The sample space of throwing two fair dice

where “the dice sum up to 4” is

Dice 1

Dice 2

1 3

2 2

3 1

4

e.g.3 () Consider the dice with “triangle”, “circle” and

“square” The sample space of throwing 2 dice of this type

isDice 1 Dice 2

TT

TCCCS

SS

TC

STC

STC

S

E: the event that at least one dice shows circle

E1: the event that the first dice shows circle

E2: the event that the second dice shows circle

P(E) = P(E1 U E2)= P(E1) + P(E2) – P(E1 E2)

P(E1) = 3/9 = 1/3 P(E2) = 3/9 = 1/3

P(E1 E2) = 1/9

= 1/3 + 1/3 – 1/9= 5/9

Assume that each possible outcome is equally likely

5



where “both top shapes are the same” is

Dice 1 Dice 2

T

CS

TCS

P(circles) = 4P(triangles)P(squares) = 9P(triangles)

P(triangles)

6



isDice 1 Dice 2

TT

TCCCS

SS

TC

STC

STC

S

1/36

1/18

1/121/181/9

1/61/121/61/4

P({TT, CC, SS}) = 1/36 + 1/9 + ¼ = 7/18

P(CC) = 1/9

7

e.g.6 (Page 9) Consider the dice with “triangle”, “circle” and


where both top shapes are the same isDice 1 Dice 2

TT

TCCCS

SS

TC

STC

STC

S

1/36

1/18

1/121/181/9

1/61/121/61/4

P({TT, CC, SS}) = 1/36 + 1/9 + ¼ = 7/18

P(TT | F ) = 1/36 / (7/18) = 1/14

P(CC | F) = 1/9 / (7/18) = 2/7

P(SS | F) = 1/4 / (7/18) = 9/14

Sum is equal to 1

Let F be the event that both top shapes are the same.

F

8

e.g.6 Consider the dice with “triangle”, “circle” and


where both top shapes are the same isDice 1 Dice 2

TT

TCCCS

SS

TC

STC

STC

S

1/36

1/18

1/121/181/9

1/61/121/61/4

P({TT, CC, SS}) = 1/36 + 1/9 + ¼ = 7/18

P(TT | F ) = 1/36 / (7/18) = 1/14

P(CC | F) = 1/9 / (7/18) = 2/7

P(SS | F) = 1/4 / (7/18) = 9/14

Let F be the event that both top shapes are the same.

F

P(F)P(CC F)

9


Dice 1

Dice 2

Sum

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

1 6 7

2 1 3

2 2 4

2 3 5

2 4 6

2 5 7

2 6 8

Dice 1

Dice 2

Sum

3 1 4

3 2 5

3 3 6

3 4 7

3 5 8

3 6 9

4 1 5

4 2 6

4 3 7

4 4 8

4 5 9

4 6 10

Dice 1

Dice 2

Sum

5 1 6

5 2 7

5 3 8

5 4 9

5 5 10

5 6 11

6 1 7

6 2 8

6 3 9

6 4 10

6 5 11

6 6 12

F: event that sum 10

P(F)= 6/36 = 1/6

10


Dice 1

Dice 2

Sum

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

1 6 7

2 1 3

2 2 4

2 3 5

2 4 6

2 5 7

2 6 8

Dice 1

Dice 2

Sum

3 1 4

3 2 5

3 3 6

3 4 7

3 5 8

3 6 9

4 1 5

4 2 6

4 3 7

4 4 8

4 5 9

4 6 10

Dice 1

Dice 2

Sum

5 1 6

5 2 7

5 3 8

5 4 9

5 5 10

5 6 11

6 1 7

6 2 8

6 3 9

6 4 10

6 5 11

6 6 12

F: event that sum 10

P(E F)= 4/36 = 1/9

E: event that sum is even

Sum=10 or 12

11

e.g.9 (Page 14)

Why is “R = (R K) U (R K)”?

Consider R

= R UniverseUniverse

R

K K

= R (K U K)

= (R K) U (R K)

12

e.g.10 (Page 14)

Given “R = (R K) U (R K)”, we know that P(R) = P(R K) + P(R

K)

Universe

R

K K

Why?

Note that R K and R K are disjoint

P(R) = P((R K) U (R K)) = P(R K) + P(R K)

13

e.g.11 (Page 16)

Suppose that we have one red dice and one green dice.

We throw these two dice. We know the following

The total sum is odd The red dice is odd

Is the green dice even or odd?We know that odd + even = odd

Thus, the green dice is even.

14

e.g.12 (Page 22)

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6

Slot 0 Slot k-1 Slot 3 Slot 5 Slot 3 Slot 4

(0, k-1, 3, 5, 3, 4)

a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)

Suppose that we have 6 items where a1 a2.

a 6-tuple

k locations/slots6 items

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

15

e.g.12

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6

Slot r Slot q Slot .. Slot .. Slot .. Slot ..

a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)


a 6-tuple


Let E be the event that a1 hashes to slot rLet F be the event that a2 hashes to slot q

(a) What is P(E)?

(b) What is P(F)?

(c) What is P(E F)?

E: the event that a1 hashes to slot rF: the event that a2 hashes to slot q(a) What is P(E)?(b) What is P(F)?(c) What is P(E F)?(d) Are E and F independent?

(d) Are E and F independent?

16

e.g.12

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)


a 6-tuple



Slot r

k choicesk choices

k choicesk choices

k choices

No. of 6-tuples where a1 hashes to slot r = k x k x k x k x k = k5

No. of 6-tuples =k x k x k x k x k x k = k6

P(E) =P(a1 hashes to slot r)= k5/k6 = 1/k

1/k

17

e.g.12

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)




1/k

Note that F is similar to E

Thus, P(F) = P(E) = 1/k

1/k

18

e.g.12

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)




1/k

1/k

a 6-tuple

Slot r

Slot qk choices

k choicesk choices

k choices

No. of 6-tuples where a1 hashes to slot r and a2 hashed to slot q = k x k x k x k = k4

No. of 6-tuples =k x k x k x k x k x k = k6

P(E F)=P(a1 hashes to slot r and a2 hashed to slot q)= k4/k6 = 1/k2

1/k2

E F: “a1 hashes to slot r and a2 hashed to slot q”

19

e.g.12

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)




1/k

1/k

a 6-tuple

1/k2

P(E) = 1/k

P(F) = 1/k

P(E F) = 1/k2

We deduce that P(E F) = P(E) x P(F)

Thus, E and F are independent.

Yes

20

e.g.13 (Page 24)

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)

Suppose that we have 6 items where a1 = a2.

a 6-tuple


Let E be the event that a1 hashes to slot rLet F be the event that a2 hashes to slot q

(a) What is P(E)?

(b) What is P(F)?

(c) What is P(E F)?


(d) Are E and F independent?

21

e.g.13

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)


a 6-tuple



1/k1/k

E F: “a1 hashes to slot r and a2 hashed to slot q”

Note that a1 = a2

We need to consider two cases.Case 1: r q

Case 2: r =q

P(E F) = 0

P(E F) = 1

0 (if r q)1 (if r = q)

22

e.g.13

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6


a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)


a 6-tuple



1/k1/k

Case 1: r q

Case 2: r =q

P(E F) = 0

P(E F) = 1

0 (if r q)1 (if r = q)

P(E) = 1/k

P(F) = 1/k

P(E) x P(F) = 1/k2

P(E) x P(F)

P(E) x P(F)

Thus, P(E F) P(E) x P(F)E and F are not independent.

No

23

e.g.14 (Page 26) Suppose that we flip a coin 5

times.We obtain the following outcome.

Suppose that we want to flip this coin again as the 6-th flip.

H T T H H

What is P(the outcome of the 6-th flip = H | HTTHH)?

We expect that this is equal to 1/2That is, it is equal to P(H)

We say that the 6-th flip is independent of theoutcome of the first 5 flips.

This is called a trial process.

This is called an independent trial process.

24

e.g.14 Suppose that we flip a coin 5

times.We obtain the following outcome.

Suppose that we want to flip this coin again as the 6-th flip.

H T T H H

What is P(the outcome of the 6-th flip = H | HTTHH)?

Let xi be the outcome of the i-th flip.

P(x6 = H | x1 = H, x2 = T, x3 = T, x4 = H, x5 = H)

The above probability can be re-written as follows according to sequenceHTTHH.

= P(x6 = H)

25

e.g.15 (Page 30)

Suppose that we flip a coin 5 times.

What is P(HTTHH)?

P(HTTHH) = P(H) x P(T) x P(T) x P(H) x P(H)= ½ x ½ x ½ x ½ x ½ x ½

= 1/256

26

e.g.16 (Page 31)

a3

a6

a2

a1

a5

a4

a2 a3 a4 a5a1a6

Slot 0 Slot k Slot 3 Slot 5 Slot 3 Slot 4

a1, a2, a3, a4, a5, a6

(l1, l2, l3, l4, l5, l6)


a 6-tuple

Slot 0

Slot 1Slot 2

Slot 3

Slot 4

Slot 5…

Slot k-1

What is P( l1 = 0, l2 = 4, l3 = 5, l4 = 1, l5= 0, l6 = 1 )?

P( l1 = 0, l2 = 4, l3 = 5, l4 = 1, l5= 0, l6 = 1 )= P(l1 = 0) x P(l2 = 4) x P(l3 = 5) x P(l4 = 1) x P(l5= 0) x P(l6 = 1)= 1/k x 1/k x 1/k x 1/k x 1/k x 1/k= 1/k6

27

e.g.17 (Page 33) Suppose that we

draw a card from a standard deck of 52 cards replace it draw another card continue for a total of ten draws.

Is the above process an independent trial process?

Yes. This is because drawing a given card at a particular time does not depend on the cards we drawn in the earlier time.

28

e.g.18 (Page 34) Suppose that we

draw a card from a standard deck of 52 cards discard it draw another card continue for a total of ten draws.

Is the above process an independent trial process?

No.

In the first draw, we have 52 cards to draw from.

In the second draw, we have only 51.

Thus, the probability for each possible outcome on the seconddraw depends on the outcome of the first draw.

29

e.g.19 (Page 36) We flip 6 coins. What is the probability that at least one coin

shows a “head”?

We learnt the principle of inclusion-and-exclusion. Let Ei be the event that coin i shows a “head”.We can compute the probability by P(E1 U E2 U E3 U E4 U E5 U E6)

This time, we can use a simple derivation by using the independent trial process.P(at least one coin shows a “head”)

= 1 – P(all coins show a “tail”)= 1 – P(F1 F2 F3 F4 F5 F6)

= 1 – P(F1) x P(F2) x P(F3) x P(F4) x P(F5) x P(F6)= 1 – ½ x ½ x ½ x ½ x ½ x ½ = 1 – 1/64 = 63/64

Let Fi be the event that coin i shows a “tail”.

1 independence supplementary notes prepared by raymond wong presented by raymond wong

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