ECE302 – Probability and Applications - Tutorial #2rrafie/ece302/Tutorial_02.pdf · 2017. 9. 28. · Question2.22 (b)FindtheprobabilitiesofeventsA;B;C;A\Bc; and A\CdeﬁnedinProblem2.2

ECE302 – Probability and ApplicationsTutorial #2

Reza Rafie

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[email protected]

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Tutorial website!www.comm.utoronto.ca/~rrafie/ece302.html

1

www.comm.utoronto.ca/~rrafie/ece302.html

Question 2.22

In Problem 2.2, a die is tossed twice and the number ofdots facing up in each toss is counted and noted in theorder of occurrence.(a) Find the probabilities of the elementary events.

2

Question 2.22

Elementary Event:

An event from a discrete samplespace that consists of a single outcome is called anelementary event.

3

Question 2.22

Elementary Event: An event from a discrete samplespace that consists of a single outcome is called anelementary event.

3

Question 2.22

S = {11, 12, . . . , 16, 21, 22, . . . , 66}

|S| = 36If the dice are unbiased:P({11}) = P({12}) = · · · = P({66}) = 1|S| =

136

How many equations? How many unknowns?Axiom II is also one equation!

4

Question 2.22

S = {11, 12, . . . , 16, 21, 22, . . . , 66}|S| = 36

If the dice are unbiased:P({11}) = P({12}) = · · · = P({66}) = 1|S| =

136


4

Question 2.22

S = {11, 12, . . . , 16, 21, 22, . . . , 66}|S| = 36If the dice are unbiased:

P({11}) = P({12}) = · · · = P({66}) = 1|S| =136


4

Question 2.22

S = {11, 12, . . . , 16, 21, 22, . . . , 66}|S| = 36If the dice are unbiased:P({11}) = P({12}) = · · · = P({66}) = 1|S| =

136


4

Question 2.22


136

How many equations? How many unknowns?

Axiom II is also one equation!

4

Question 2.22


136


4

Question 2.22

(b) Find the probabilities of events A,B, C,A ∩ Bc, andA ∩ C defined in Problem 2.2.A: event that “number of dots in first toss is not less thannumber of dots in second toss”B: event that “number of dots in first toss is 6.”C: event that “number of dots in dice differs by 2.”

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Question 2.22

If the dice are unbiased:

P(A) = |A||S|A = {11, 21, 22, 31, . . . , 66}|A| = 21=⇒ P(A) = 2136

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Question 2.22

If the dice are unbiased:P(A) = |A||S|

A = {11, 21, 22, 31, . . . , 66}|A| = 21=⇒ P(A) = 2136

6

Question 2.22

If the dice are unbiased:P(A) = |A||S|A = {11, 21, 22, 31, . . . , 66}

|A| = 21=⇒ P(A) = 2136

6

Question 2.22

If the dice are unbiased:P(A) = |A||S|A = {11, 21, 22, 31, . . . , 66}|A| =

21=⇒ P(A) = 2136

6

Question 2.22

If the dice are unbiased:P(A) = |A||S|A = {11, 21, 22, 31, . . . , 66}|A| = 21

=⇒ P(A) = 2136

6

Question 2.22

If the dice are unbiased:P(A) = |A||S|A = {11, 21, 22, 31, . . . , 66}|A| = 21=⇒ P(A) = 2136

6

Question 2.22

P(B) = |B||S|

B = {61, 62, . . . , 66}|B| = 6=⇒ P(B) = 636

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Question 2.22

P(B) = |B||S|B = {61, 62, . . . , 66}

|B| = 6=⇒ P(B) = 636

7

Question 2.22

P(B) = |B||S|B = {61, 62, . . . , 66}|B| =

6=⇒ P(B) = 636

7

Question 2.22

P(B) = |B||S|B = {61, 62, . . . , 66}|B| = 6

=⇒ P(B) = 636

7

Question 2.22

P(B) = |B||S|B = {61, 62, . . . , 66}|B| = 6=⇒ P(B) = 636

7

Question 2.22

P(C) = |C||S|

C = {13, 24, 35, 46, 31, 42, 53, 64}|C| = 8=⇒ P(C) = 836

8

Question 2.22

P(C) = |C||S|C =

{13, 24, 35, 46, 31, 42, 53, 64}|C| = 8=⇒ P(C) = 836

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Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46

, 31, 42, 53, 64}|C| = 8=⇒ P(C) = 836

8

Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46, 31

, 42, 53, 64}|C| = 8=⇒ P(C) = 836

8

Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46, 31, 42, 53, 64}

|C| = 8=⇒ P(C) = 836

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Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46, 31, 42, 53, 64}|C| =

8=⇒ P(C) = 836

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Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46, 31, 42, 53, 64}|C| = 8

=⇒ P(C) = 836

8

Question 2.22

P(C) = |C||S|C = {13, 24, 35, 46, 31, 42, 53, 64}|C| = 8=⇒ P(C) = 836

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|

A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}=⇒ |A ∩ Bc| = 21− 6 = 15=⇒ P(A ∩ Bc) = 1536Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}

=⇒ |A ∩ Bc| = 21− 6 = 15=⇒ P(A ∩ Bc) = 1536Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}=⇒ |A ∩ Bc| =

21− 6 = 15=⇒ P(A ∩ Bc) = 1536Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}=⇒ |A ∩ Bc| = 21− 6 = 15

=⇒ P(A ∩ Bc) = 1536Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}=⇒ |A ∩ Bc| = 21− 6 = 15=⇒ P(A ∩ Bc) = 1536

Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ Bc) = |A∩Bc|

|S|A ∩ Bc = A− B = {11, 21, 22, 31, . . . , 66} − {61, 62, . . . , 66}=⇒ |A ∩ Bc| = 21− 6 = 15=⇒ P(A ∩ Bc) = 1536Theorem:Let A and B be two finite sets. Then |A− B| = |A| − |B| ifand only if (iff ) B ⊂ A.

Try to prove it!

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Question 2.22

P(A ∩ C) = |A∩C||S|

A ∩ C = {31, 42, 53, 64}|A ∩ C| = 4=⇒ P(A ∩ C) = 436

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Question 2.22

P(A ∩ C) = |A∩C||S|A ∩ C = {31, 42, 53, 64}

|A ∩ C| = 4=⇒ P(A ∩ C) = 436

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Question 2.22

P(A ∩ C) = |A∩C||S|A ∩ C = {31, 42, 53, 64}|A ∩ C| = 4

=⇒ P(A ∩ C) = 436

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Question 2.22

P(A ∩ C) = |A∩C||S|A ∩ C = {31, 42, 53, 64}|A ∩ C| = 4=⇒ P(A ∩ C) = 436

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Question 2.23

A random experiment has sample space S = {a,b, s,d}.Suppose that P({c,d}) = 3/8, P({b, c}) = 6/8 andP({d}) = 1/8. Use the axioms of probability to find theprobabilities of the elementary events.

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Question 2.23

P({a}) =?P({b}) =?P({c}) =?P({d}) =?

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Question 2.23

3/8 = P({c,d})

= P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)

6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c})

= P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)

1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)

How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?

Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II:

P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 1

1 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)

From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8

(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

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Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8

(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

13

Question 2.23

3/8 = P({c,d}) = P({c}) + P({d}) (i)6/8 = P({b, c}) = P({c}) + P({b}) (ii)1/8 = P({d}) (iii)How many equation? How many unknowns?Axiom II: P(S) = 11 = P({a}) + P({b}) + P({c}) + P({d}) (iv)From (i) and (iii) =⇒ P({c}) = 2/8(ii) =⇒ P({b}) = 4/8(iv) =⇒ P({a}) = 1− 1/8− 2/8− 4/8 = 1/8

13

Question 2.30

Use Corollary 7 to prove the following:a) P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

14

Question 2.30

Use Corollary 5 instead!

P(X ∪ Y) = P(X) + P(Y)− P(X ∩ Y)Axiom I: P(X ∩ Y) ≥ 0=⇒ P(X ∪ Y) ≤ P(X) + P(Y)

15

Question 2.30

Use Corollary 5 instead!P(X ∪ Y) = P(X) + P(Y)− P(X ∩ Y)

Axiom I: P(X ∩ Y) ≥ 0=⇒ P(X ∪ Y) ≤ P(X) + P(Y)

15

Question 2.30

Use Corollary 5 instead!P(X ∪ Y) = P(X) + P(Y)− P(X ∩ Y)Axiom I: P(X ∩ Y) ≥ 0

=⇒ P(X ∪ Y) ≤ P(X) + P(Y)

15

Question 2.30

Use Corollary 5 instead!P(X ∪ Y) = P(X) + P(Y)− P(X ∩ Y)Axiom I: P(X ∩ Y) ≥ 0=⇒ P(X ∪ Y) ≤ P(X) + P(Y)

15

Question 2.30

P(X ∪ Y) ≤ P(X) + P(Y) (i)

Let X = A ∪ B and Y = C(i) =⇒ P(A ∪ B ∪ C) ≤ P(A ∪ B) + P(C) (ii)Now let X = A and Y = B in (i)(i), (ii) =⇒ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

16

Question 2.30

P(X ∪ Y) ≤ P(X) + P(Y) (i)Let X = A ∪ B and Y = C

(i) =⇒ P(A ∪ B ∪ C) ≤ P(A ∪ B) + P(C) (ii)Now let X = A and Y = B in (i)(i), (ii) =⇒ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

16

Question 2.30

P(X ∪ Y) ≤ P(X) + P(Y) (i)Let X = A ∪ B and Y = C(i) =⇒ P(A ∪ B ∪ C) ≤ P(A ∪ B) + P(C) (ii)

Now let X = A and Y = B in (i)(i), (ii) =⇒ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

16

Question 2.30

P(X ∪ Y) ≤ P(X) + P(Y) (i)Let X = A ∪ B and Y = C(i) =⇒ P(A ∪ B ∪ C) ≤ P(A ∪ B) + P(C) (ii)Now let X = A and Y = B in (i)

(i), (ii) =⇒ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

16

Question 2.30

P(X ∪ Y) ≤ P(X) + P(Y) (i)Let X = A ∪ B and Y = C(i) =⇒ P(A ∪ B ∪ C) ≤ P(A ∪ B) + P(C) (ii)Now let X = A and Y = B in (i)(i), (ii) =⇒ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)

16

Question 2.30

b) P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)

Induction!

17

Question 2.30

b) P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)How do you approach such a problem?

Induction!

17

Question 2.30

b) P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)Induction!

17

Question 2.30

Principle of Mathematical Induction:Let S(n) denote an open mathematical statement thatinvolves one or more occurrences of variable n. Also,let n0 ≤ n1 be natural numbers.

• (base) If S(n0), S(n0 + 1), S(n0 + 2), . . . , S(n1 − 1) andS(n1) are true; and

• (step) if whenever S(n0), S(n0 + 1), . . . , S(k− 1) andS(k) are true for some (particular but arbitrarychosen) natural number k ≥ n1, then the statementS(k+ 1) is also true;

then S(n) is true for all natural numbers n ≥ n0.

18

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)

(base): n0 = n1 = 2S(n0): P(A1 ∪ A2) ≤ P(A1) + P(A2)This is true by equation (i) in part (a).

19

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(base): n0 = n1 = 2

S(n0): P(A1 ∪ A2) ≤ P(A1) + P(A2)This is true by equation (i) in part (a).

19

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(base): n0 = n1 = 2S(n0): P(A1 ∪ A2) ≤ P(A1) + P(A2)

This is true by equation (i) in part (a).

19

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(base): n0 = n1 = 2S(n0): P(A1 ∪ A2) ≤ P(A1) + P(A2)This is true by equation (i) in part (a).

19

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(step): Let S(n) be true for n = 2, 3, . . . , k.

We prove S(k+ 1)S(k+ 1): P

(∪k+1i=1 Ai

)≤∑k+1

i=1 P(Ai)

S(2) =⇒ P(k+1∪i=1

Ai

)≤ P

( k∪i=1

Ai

)+ P(Ak+1)

S(k)≤

k∑i=1

P(Ai) + P(Ak+1)

=k+1∑i=1

P(Ai)

=⇒ S(k+ 1)

20

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(step): Let S(n) be true for n = 2, 3, . . . , k.We prove S(k+ 1)

S(k+ 1): P(∪k+1

i=1 Ai)≤∑k+1

i=1 P(Ai)

S(2) =⇒ P(k+1∪i=1

Ai

)≤ P

( k∪i=1

Ai

)+ P(Ak+1)

S(k)≤

k∑i=1

P(Ai) + P(Ak+1)

=k+1∑i=1

P(Ai)

=⇒ S(k+ 1)

20

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(step): Let S(n) be true for n = 2, 3, . . . , k.We prove S(k+ 1)S(k+ 1): P

(∪k+1i=1 Ai

)≤∑k+1

i=1 P(Ai)

S(2) =⇒ P(k+1∪i=1

Ai

)≤ P

( k∪i=1

Ai

)+ P(Ak+1)

S(k)≤

k∑i=1

P(Ai) + P(Ak+1)

=k+1∑i=1

P(Ai)

=⇒ S(k+ 1)

20

Question 2.30

S(n): P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)(step): Let S(n) be true for n = 2, 3, . . . , k.We prove S(k+ 1)S(k+ 1): P

(∪k+1i=1 Ai

)≤∑k+1

i=1 P(Ai)

S(2) =⇒ P(k+1∪i=1

Ai

)≤ P

( k∪i=1

Ai

)+ P(Ak+1)

S(k)≤

k∑i=1

P(Ai) + P(Ak+1)

=k+1∑i=1

P(Ai)

=⇒ S(k+ 1) 20

Question 2.30

The proof is not complete yet!

By mathematical induction, the statement is true for allnatural numbers n ≥ 2 (and trivial for n = 1).

21

Question 2.30

The proof is not complete yet!By mathematical induction, the statement is true for allnatural numbers n ≥ 2 (and trivial for n = 1).

21

Question 2.30

b) P(∪n

i=1 Ai)≤∑n

i=1 P(Ai)This is called the union bound.

22

Question 2.30

c) P(∩n

i=1 Ai)≥ 1−

∑ni=1 P(Aci )

23

Question 2.30

P( n∩i=1

Ai

)Crollary 1= 1− P

([ n∩i=1

Ai

]c)

DeMorgan’s law= 1− P

( n∪i=1

Aci

)Part (b)≥ 1−

n∑i=1

P(Aci )

24

Question 2.30

P( n∩i=1

Ai

)Crollary 1= 1− P

([ n∩i=1

Ai

]c)DeMorgan’s law

= 1− P( n∪i=1

Aci

)

Part (b)≥ 1−

n∑i=1

P(Aci )

24

Question 2.30

P( n∩i=1

Ai

)Crollary 1= 1− P

([ n∩i=1

Ai

]c)DeMorgan’s law

= 1− P( n∪i=1

Aci

)Part (b)≥ 1−

n∑i=1

P(Aci )

24

Question 2.32

A die is tossed and the number of dots facing up isnoted. Find the probability of the elementary events iffaces with an even number of dots are twice as likely tocome up as faces with an odd number.

25

Question 2.32

S = {1, 2, 3, 4, 5, 6}

P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4)

= 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)

x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)

P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1

= x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x

= 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x

=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9

P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.32

S = {1, 2, 3, 4, 5, 6}P(2) = P(4) = P(4) = 2P(1) = 2P(3) = 2P(5)x , P(1)P(S) = 1 = x+ 2x+ x+ 2x+ x+ 2x = 9x=⇒ x = 1/9P(2) = P(4) = P(4) = 2/9P(1) = P(3) = P(5) = 1/9

26

Question 2.35

A number x is selected at random in the interval [−1, 2].Numbers from the subinterval [0, 2] occur half asfrequently as those from [−1, 0). Find the probabilityassignment for an interval completely within [−1, 0);completely within [0, 2]; and partly in each of the aboveintervals.

27

Question 2.35

P([0, 2]) = 0.5P([−1, 0))

P([0, 2]) + P([−1, 0)) = 1=⇒ P([0, 2]) = 13 ,P([−1, 0)) =

23

28

Question 2.35

P([0, 2]) = 0.5P([−1, 0))P([0, 2]) + P([−1, 0)) = 1

=⇒ P([0, 2]) = 13 ,P([−1, 0)) =23

28

Question 2.35

P([0, 2]) = 0.5P([−1, 0))P([0, 2]) + P([−1, 0)) = 1=⇒ P([0, 2]) = 13 ,P([−1, 0)) =

23

28

Question 2.35

If I = [a,b] is an interval contained in [0, 2], we expect tohave P(I) ∝ b− a.

That is, for all 0 ≤ a ≤ b ≤ 2 we have

P([a,b]) = C0 × [b− a]

Let b = 2,a = 0=⇒ P([0, 2]) = C0 × (2− 0)We already know that P([0, 2]) = 13 =⇒ C0 =

16 .

=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35

If I = [a,b] is an interval contained in [0, 2], we expect tohave P(I) ∝ b− a.That is, for all 0 ≤ a ≤ b ≤ 2 we have

P([a,b]) = C0 × [b− a]


16 .

=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35


P([a,b]) = C0 × [b− a]

Let b = 2,a = 0

=⇒ P([0, 2]) = C0 × (2− 0)We already know that P([0, 2]) = 13 =⇒ C0 =

16 .

=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35


P([a,b]) = C0 × [b− a]

Let b = 2,a = 0=⇒ P([0, 2]) = C0 × (2− 0)

We already know that P([0, 2]) = 13 =⇒ C0 =16 .

=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35


P([a,b]) = C0 × [b− a]

Let b = 2,a = 0=⇒ P([0, 2]) = C0 × (2− 0)We already know that P([0, 2]) = 13

=⇒ C0 = 16 .=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35


P([a,b]) = C0 × [b− a]


16 .

=⇒ if I = [a,b] ⊂ [0, 2] then P(I) = b−a6

29

Question 2.35

Note: A singular point does not carry probability

P([a,a]) = C× (a− a) = 0

Therefore, we do not discriminate between(a,b), (a,b], [a,b) and [a,b]. This argument is notmathematically precise yet, but makes intuitive sense.

30

Question 2.35


P([a,a]) = C× (a− a) = 0

Therefore, we do not discriminate between(a,b), (a,b], [a,b) and [a,b].

This argument is notmathematically precise yet, but makes intuitive sense.

30

Question 2.35


P([a,a]) = C× (a− a) = 0

Therefore, we do not discriminate between(a,b), (a,b], [a,b) and [a,b]. This argument is notmathematically precise yet, but makes intuitive sense.

30

Question 2.35

Similarly, if I = [a,b) is an interval contained in [−1, 0),we expect to have P(I) ∝ b− a.

That is, for all 0 ≤ a ≤ b ≤ 2 we have

P([a,b]) = C1 × [b− a]

Let b = −1,a = 0=⇒ P([−1, 0)) = C1 × (0− (−1))We already know that P([−1, 0)) = 23 =⇒ C1 =

23 .

=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35

Similarly, if I = [a,b) is an interval contained in [−1, 0),we expect to have P(I) ∝ b− a.That is, for all 0 ≤ a ≤ b ≤ 2 we have

P([a,b]) = C1 × [b− a]


23 .

=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35


P([a,b]) = C1 × [b− a]

Let b = −1,a = 0

=⇒ P([−1, 0)) = C1 × (0− (−1))We already know that P([−1, 0)) = 23 =⇒ C1 =

23 .

=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35


P([a,b]) = C1 × [b− a]

Let b = −1,a = 0=⇒ P([−1, 0)) = C1 × (0− (−1))

We already know that P([−1, 0)) = 23 =⇒ C1 =23 .

=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35


P([a,b]) = C1 × [b− a]

Let b = −1,a = 0=⇒ P([−1, 0)) = C1 × (0− (−1))We already know that P([−1, 0)) = 23

=⇒ C1 = 23 .=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35


P([a,b]) = C1 × [b− a]


23 .

=⇒ if I = [a,b] ⊂ [−1, 0) then P(I) = 2(b−a)3

31

Question 2.35

If I = [a,b] is an interval contained in [−1, 2] with a < 0and b ≥ 0, then we have

I = [a, 0)︸︷︷︸I1

∪ [0,b]︸︷︷︸I2

I1 ∩ I2 = 0

Axiom III=⇒ P(I) = P(I1) + P(I2)

=⇒ P(I) = C0 × (b− 0) + C1 × (0− a) =b6 −

2a3

Note: Break down the event into a collection of disjointevents that can be dealt with easier!

32

Question 2.35


I = [a, 0) ∪ [0,b]

I = [a, 0)︸︷︷︸I1

∪ [0,b]︸︷︷︸I2

I1 ∩ I2 = 0


=⇒ P(I) = C0 × (b− 0) + C1 × (0− a) =b6 −

2a3


32

Question 2.35


I = [a, 0)︸︷︷︸I1

∪ [0,b]︸︷︷︸I2

I1 ∩ I2 = 0


=⇒ P(I) = C0 × (b− 0) + C1 × (0− a) =b6 −

2a3


32

Question 2.35


I = [a, 0)︸︷︷︸I1

∪ [0,b]︸︷︷︸I2

I1 ∩ I2 = 0


=⇒ P(I) = C0 × (b− 0) + C1 × (0− a) =b6 −

2a3

Note: Break down the event into a collection of disjointevents that can be dealt with easier! 32

Example 2.16

An urn contains five balls numbered 1 to 5. Suppose weselect two balls in succession without replacement. Howmany distinct ordered pairs are possible?

33

Example 2.16

The Rule of Product:If a procedure can be broken down into first andsecond stages, if there are m possible outcomes for thefirst stage and if, for each of those outcomes, there aren possible outcomes for the second stage, then the toalprocedure can be carried out, in the designated order,in mn ways.

Generalize by induction!

34

Example 2.16

There are 5 possibilities for the first ball

,for each outcome of the first ball, there are 4 possibilitiesfor the second ball.The number of ordered pairs is 5× 4 = 20.

35

Example 2.16

There are 5 possibilities for the first ball,for each outcome of the first ball, there are 4 possibilitiesfor the second ball.

The number of ordered pairs is 5× 4 = 20.

35

Example 2.16

There are 5 possibilities for the first ball,for each outcome of the first ball, there are 4 possibilitiesfor the second ball.The number of ordered pairs is 5× 4 = 20.

35

Example 2.16

What is the probability that the first ball has a numberlarger than that of the second ball?

36

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space).

It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.Therefore, exactly half of the outcomes satisfy therequired condition.Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.Check your textbook for a detailed solution!P(A) = |A||S|

37

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space). It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.

Therefore, exactly half of the outcomes satisfy therequired condition.Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.Check your textbook for a detailed solution!P(A) = |A||S|

37

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space). It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.Therefore, exactly half of the outcomes satisfy therequired condition.

Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.Check your textbook for a detailed solution!P(A) = |A||S|

37

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space). It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.Therefore, exactly half of the outcomes satisfy therequired condition.Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.

Check your textbook for a detailed solution!P(A) = |A||S|

37

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space). It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.Therefore, exactly half of the outcomes satisfy therequired condition.Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.Check your textbook for a detailed solution!

P(A) = |A||S|

37

Example 2.16

Let (a,b) ∈ S be an arbitrary outcome (here S is thesample space). It is obvious that (b,a) ̸= (a,b) and(b,a) ∈ S.Therefore, exactly half of the outcomes satisfy therequired condition.Thus, the probability of the event that the first ball has anumber larger than that of the second ball is 0.5.Check your textbook for a detailed solution!P(A) = |A||S|

37

Question 2.52

A dinner party is attended by four men and four women.How many unique ways can the eight people sit aroundthe table?

38

Question 2.52

If we want them to stay in a line, we could use The Ruleof Product and get 8! = 8× 7× · · · × 1 permutations.

But now every circular arrangement corresponds to 8different linear arrangements.Thus, there are 8!8 = 7! ways in total.

39

Question 2.52

If we want them to stay in a line, we could use The Ruleof Product and get 8! = 8× 7× · · · × 1 permutations.But now every circular arrangement corresponds to 8different linear arrangements.

Thus, there are 8!8 = 7! ways in total.

39

Question 2.52

If we want them to stay in a line, we could use The Ruleof Product and get 8! = 8× 7× · · · × 1 permutations.But now every circular arrangement corresponds to 8different linear arrangements.Thus, there are 8!8 = 7! ways in total.

39

Question 2.52

How many unique ways can the people sit around thetable with men and women alternating seats?

40

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.

Now rotate clockwise and fill in the blank positions.Next to Mr. X, there are 4 women that can sit (4possibilities)After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)After that, there are 3 women that can sit in the nextposition (3 possibilities)...

41

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.Now rotate clockwise and fill in the blank positions.

Next to Mr. X, there are 4 women that can sit (4possibilities)After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)After that, there are 3 women that can sit in the nextposition (3 possibilities)...

41

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.Now rotate clockwise and fill in the blank positions.Next to Mr. X, there are 4 women that can sit (4possibilities)

After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)After that, there are 3 women that can sit in the nextposition (3 possibilities)...

41

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.Now rotate clockwise and fill in the blank positions.Next to Mr. X, there are 4 women that can sit (4possibilities)After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)

After that, there are 3 women that can sit in the nextposition (3 possibilities)...

41

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.Now rotate clockwise and fill in the blank positions.Next to Mr. X, there are 4 women that can sit (4possibilities)After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)After that, there are 3 women that can sit in the nextposition (3 possibilities)

...

41

Question 2.52

Put Mr. X at a fixed position. This position for Mr. X canbe obtained from any given arrangement by circularshifting of the table.Now rotate clockwise and fill in the blank positions.Next to Mr. X, there are 4 women that can sit (4possibilities)After the first woman is set, there are 3 men that can sitin the next position (3 possibilities)After that, there are 3 women that can sit in the nextposition (3 possibilities)...

41

Question 2.52

In total, there will be 4× 3× 3× 2× 2× 1× 1 ways!

42

Question 2.52

Other solution:

Try to arrange them on a line, and use then use theargument that “every circular arrangement correspondsto 8 different linear arrangements.”You should get 2×4!×4!8

43

Question 2.52

Other solution:Try to arrange them on a line, and use then use theargument that “every circular arrangement correspondsto 8 different linear arrangements.”

You should get 2×4!×4!8

43

Question 2.52

Other solution:Try to arrange them on a line, and use then use theargument that “every circular arrangement correspondsto 8 different linear arrangements.”You should get 2×4!×4!8

43

Question 2.56

A lot of 50 items has 40 good items and 10 bad items.a) Suppose we test five samples from the lot, withreplacement. Let X be the number of defective items inthe sample. Find P(X = k).

44

Question 2.56

Note: Instead of P({outcumes such that X = k}), for thesake of brevity, we usually write P(X = k).

45

Question 2.56

Probability of having a defective item: pd =

10/50 = 1/5Probability of having a good item: pg = 40/50 = 4/5

46

Question 2.56

Probability of having a defective item: pd = 10/50 = 1/5

Probability of having a good item: pg = 40/50 = 4/5

46

Question 2.56

Probability of having a defective item: pd = 10/50 = 1/5Probability of having a good item: pg = 40/50 = 4/5

46

Question 2.56


P(X = k) =(5k

)× pkd × p5−kg

46

Question 2.56


P(X = k) =(5k

)× pkd︸︷︷︸

probability of k defective items

×p5−kg

46

Question 2.56


P(X = k) =(5k

)× pkd︸︷︷︸probability of k defective items

× p5−kg︸︷︷︸probability of 5−k good items

46

Question 2.56


P(X = k) =(5k

)×pkd × p5−kg︸︷︷︸

probability of a particular sequence with k defective items

46

Question 2.56


P(X = k) =(5k

)︸︷︷︸

number of sequences with k defective items

×pkd × p5−kg

46

Question 2.56

A lot of 50 items has 40 good items and 10 bad items.b) Suppose we test five samples from the lot, withoutreplacement. Let Y be the number of defective items inthe sample. Find P(Y = k).

47

Question 2.56

Number of ways we can pick 5 items out of 50:

|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)

(505 )Can you think of a direct method?

48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46

= 50!45!Number of such sequences with k defective items:(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!

Number of such sequences with k defective items:(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:

(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:(10k

)︸︷︷︸Number of ways we can pick k out of 10 defective items,without ordering them yet

(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:(10k)×(405− k

)︸︷︷︸

Number of ways we can pick 5− k out of 40 good items,without ordering them yet

(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:(10k)×( 405−k)× 5!︸︷︷︸

Number of ways we can order these 5 samples

(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56

Number of ways we can pick 5 items out of 50:|S| = 50× 49× · · · × 46 = 50!45!Number of such sequences with k defective items:(10k)( 40

5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.56


5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)

(505 )

Can you think of a direct method?

48

Question 2.56


5−k)5!

P(Y = k) = (10k)(

405−k)5!50!45!

=(10k)(

405−k)


48

Question 2.62

A die is tossed twice and the number of dots facing up iscounted and noted in the order of occurrence. Let A bethe event “number of dots in first toss is not less thannumber of dots in second toss,” and let B be the event“number of dots in first toss is 6.”Find P[A|B] and P[B|A].

49

Question 2.62

The conditional probability is defined as

P[A|B] = P[A ∩ B]P[B]

B = {61, 62, . . . , 66} = A ∩ B

P[A|B] = 1

50

Question 2.62


P[B|A] = P[A ∩ B]P[A]

A ∩ B = {61, 62, . . . , 66}

A = {11, 21, 22, 31, . . . , 66} =⇒ |A| = 21

P[B|A] = 621 =27

51

Question 2.62


P[B|A] = P[A ∩ B]P[A]

A ∩ B = {61, 62, . . . , 66}

A = {11, 21, 22, 31, . . . , 66} =⇒ |A| =

21

P[B|A] = 621 =27

51

Question 2.62


P[B|A] = P[A ∩ B]P[A]

A ∩ B = {61, 62, . . . , 66}

A = {11, 21, 22, 31, . . . , 66} =⇒ |A| = 21

P[B|A] = 621 =27

51

Question 2.69

A number x is selected at random in the interval [−1, 2].Let the events A = {x < 0}, B{|x− 0.5| < 0.5}, andC = {x > 0.75}. Find P[A|B], P[B|C], P[A|Cc], P[B|Cc].

52

Question 2.69

S = [−1, 2], |S| = 3A = {x < 0}

B{|x− 0.5| < 0.5}C = {x > 0.75}

53

Question 2.69

+∞−∞ 0

2−1

S

A

B1

C0.75

54

Question 2.69

+∞−∞ 0 2−1S

A

B1

C0.75

54

Question 2.69

00

A

B1

P[A|B] = P[A∩B]P[B] =

P(∅)P(B) = 0

55

Question 2.69

00

A

B1

P[A|B] = P[A∩B]P[B] =P(∅)

P(B) = 0

55

Question 2.69

00

A

B1

P[A|B] = P[A∩B]P[B] =P(∅)P(B)

= 0

55

Question 2.69

00

A

B1

P[A|B] = P[A∩B]P[B] =P(∅)P(B) = 0

55

Question 2.69

0

B1

C0.75 2

P[B|C] = P[B∩C]P[C] =

P((0.75,1))P((0.75,2]) =

0.251.25 =

15

56

Question 2.69

0

B1

C0.75 2

P[B|C] = P[B∩C]P[C] =P((0.75,1))

P((0.75,2]) =0.251.25 =

15

56

Question 2.69

0

B1

C0.75 2

P[B|C] = P[B∩C]P[C] =P((0.75,1))P((0.75,2])

= 0.251.25 =15

56

Question 2.69

0

B1

C0.75 2

P[B|C] = P[B∩C]P[C] =P((0.75,1))P((0.75,2]) =

0.251.25 =

15

56

Question 2.69

0

B1

C0.75 2

Cc0.75−1

P[B|Cc] = P[B∩Cc]

P[Cc] =

P((0,0.75])P([−1,0.75]) =

0.751.75 =

37

57

Question 2.69

0

B1

C0.75 2

Cc0.75−1

P[B|Cc] = P[B∩Cc]

P[Cc] =P((0,0.75])

P([−1,0.75]) =0.751.75 =

37

57

Question 2.69

0

B1

C0.75 2

Cc0.75−1

P[B|Cc] = P[B∩Cc]

P[Cc] =P((0,0.75])P([−1,0.75])

= 0.751.75 =37

57

Question 2.69

0

B1

C0.75 2

Cc0.75−1

P[B|Cc] = P[B∩Cc]

P[Cc] =P((0,0.75])P([−1,0.75]) =

0.751.75 =

37

57

Question 2.69

0−1

A

Cc0.75−1

P[A|Cc] = P[A∩Cc]

P[Cc] =

P([−1,0))P([−1,0.75]) =

11.75 =

47

58

Question 2.69

0−1

A

Cc0.75−1

P[A|Cc] = P[A∩Cc]

P[Cc] =P([−1,0))

P([−1,0.75]) =11.75 =

47

58

Question 2.69

0−1

A

Cc0.75−1

P[A|Cc] = P[A∩Cc]

P[Cc] =P([−1,0))P([−1,0.75])

= 11.75 =47

58

Question 2.69

0−1

A

Cc0.75−1

P[A|Cc] = P[A∩Cc]

P[Cc] =P([−1,0))P([−1,0.75]) =

11.75 =

47

58

Question 2.77

A nonsymmetric binary communications channel isshown in the figure below. Assume the input is 0 withprobability p and 1 with probability 1− p.

(a) Find the probability that the output is 0.1− ϵ1

1− ϵ2

ϵ1ϵ2

0 0

1 1

59

Question 2.77

Input: XOutput: Y

60

Question 2.77

1− ϵ1

1− ϵ2

ϵ1ϵ2

0 0

1 1

P[{Y = 0}] = P[{Y = 0} ∩ S]= P[{Y = 0} ∩ ({X = 0} ∪ {X = 1})]= P [({Y = 0} ∩ {X = 0}) ∪ ({Y = 0} ∩ {X = 1})]= P [({Y = 0} ∩ {X = 0})] + P [({Y = 0} ∩ {X = 1})]

61

Question 2.771− ϵ1

1− ϵ2

ϵ1ϵ2

0 0

1 1

P [({Y = 0} ∩ {X = 0})] = P [({Y = 0}|{X = 0})]× P [{X = 0}]= (1− ϵ1)× p

P [({Y = 0} ∩ {x = 1})] = P [({Y = 0}|{X = 1})]× P [{X = 1}]= ϵ2 × (1− p)

P [{Y = 0}] = (1− ϵ1)× p+ ϵ2 × (1− p)

62

Question 2.77

Remark:To evaluate P(Y), one can partition S into disjointsubsets X1, X2, . . . , Xn and use the law of totalprobability:

P(Y) =i=n∑i=1

P(Y|Xi)× P(Xi)

63

Question 2.77

(b) Find the probability that the input was 0 given thatthe output is 1. Find the probability that the input is 1given that the output is 1. Which input is more probable?

64

Question 2.77

Probability that the input was 0 given that the output is 1= P[X = 0|Y = 1]

P[X = 0|Y = 1] = P({X = 0} ∩ {Y = 1})P(Y = 1) (1)

65

Question 2.77

We have P(Y|X) and P(Y = 0), so we want to writeeverything in terms of P(Y|X) and P(Y = 0)

Denominator:P(Y = 1) = 1− P(Y = 0) = 1− (1− ϵ1)× p− ϵ2 × (1− p)Numerator:By definition of conditional probability,P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0})× P(X = 0) = ϵ1 × p

66

Question 2.77

We have P(Y|X) and P(Y = 0), so we want to writeeverything in terms of P(Y|X) and P(Y = 0)Denominator:P(Y = 1) = 1− P(Y = 0)

= 1− (1− ϵ1)× p− ϵ2 × (1− p)Numerator:By definition of conditional probability,P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0})× P(X = 0) = ϵ1 × p

66

Question 2.77

We have P(Y|X) and P(Y = 0), so we want to writeeverything in terms of P(Y|X) and P(Y = 0)Denominator:P(Y = 1) = 1− P(Y = 0) = 1− (1− ϵ1)× p− ϵ2 × (1− p)

Numerator:By definition of conditional probability,P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0})× P(X = 0) = ϵ1 × p

66

Question 2.77

We have P(Y|X) and P(Y = 0), so we want to writeeverything in terms of P(Y|X) and P(Y = 0)Denominator:P(Y = 1) = 1− P(Y = 0) = 1− (1− ϵ1)× p− ϵ2 × (1− p)Numerator:By definition of conditional probability,P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0})× P(X = 0)

= ϵ1 × p

66

Question 2.77

We have P(Y|X) and P(Y = 0), so we want to writeeverything in terms of P(Y|X) and P(Y = 0)Denominator:P(Y = 1) = 1− P(Y = 0) = 1− (1− ϵ1)× p− ϵ2 × (1− p)Numerator:By definition of conditional probability,P({X = 0} ∩ {Y = 1}) = P({Y = 1}|{X = 0})× P(X = 0) = ϵ1 × p

66

Question 2.77

Probability that the input was 0 given that the output is 1= P[X = 0|Y = 1]

P[X = 0|Y = 1] = P({X = 0} ∩ {Y = 1})P(Y = 1) (2)

=ϵ1 × p

1− (1− ϵ1)× p− ϵ2 × (1− p)(3)

67

Question 2.77

Similarly, probability that the input was 1 given that theoutput is 1 = P[X = 1|Y = 1]

P[X = 1|Y = 1] = P({X = 1} ∩ {Y = 1})P(Y = 1) (4)

=(1− ϵ2)× (1− p)

1− (1− ϵ1)× p− ϵ2 × (1− p)(5)

68

Question 2.77

Which input is more probable?

Same denominators, onlyneed to compare numerators

?ϵ1 × p R (1− ϵ2)× (1− p)

69

Question 2.77

Which input is more probable? Same denominators, onlyneed to compare numerators

?ϵ1 × p R (1− ϵ2)× (1− p)

69

Question 2.80

A computer manufacturer uses chips from three sources.Chips from sources A, B, and C are defective withprobabilities 0.005, 0.001, and 0.01, respectively. If arandomly selected chip is found to be defective, find theprobability that the manufacturer was A; that themanufacturer was C. Assume that the proportions ofchips from A, B, and C are 0.5, 0.1, and 0.4, respectively.

70

Question 2.80

What do we know?

P(A) = 0.5P(B) = 0.1P(C) = 0.4P(Defective|A) = 0.005P(Defective|B) = 0.001P(Defective|C) = 0.01

71

Question 2.80

What do we know?P(A) = 0.5P(B) = 0.1P(C) = 0.4

P(Defective|A) = 0.005P(Defective|B) = 0.001P(Defective|C) = 0.01

71

Question 2.80

What do we know?P(A) = 0.5P(B) = 0.1P(C) = 0.4P(Defective|A) = 0.005

P(Defective|B) = 0.001P(Defective|C) = 0.01

71

Question 2.80

What do we know?P(A) = 0.5P(B) = 0.1P(C) = 0.4P(Defective|A) = 0.005P(Defective|B) = 0.001P(Defective|C) = 0.01

71

Question 2.80

Remark:Bayes’ rule is a way to relate P(X|Y) to P(Y|X):

P(X|Y) = P(Y|X)P(X)P(Y)

Most of the time, the law of total probability is used tofind the denominator in Bayes’ formula.

72

Question 2.80

P(A|Defective) = P(Defective|A)P(A)P(Defective)

Denominator:


P(Defective) = 0.005× 0.5+ 0.001× 0.1+ 0.01× 0.4 = 0.0066

73

Question 2.80

P(A|Defective) = P(Defective|A)P(A)P(Defective)Denominator:


P(Defective) = 0.005× 0.5+ 0.001× 0.1+ 0.01× 0.4 = 0.0066

73

Question 2.80

P(A|Defective) =P(Defective|A)P(A)P(Defective)0.005× 0.50.0066 =

2566

74

Question 2.80

Similarly,

P(C|Defective) =P(Defective|C)P(C)P(Defective)0.01× 0.40.0066 =

4066

75

Questions?

75

Documents

ECE302 – Probability and Applications - Tutorial #2rrafie/ece302/Tutorial_02.pdf · 2017. 9. 28. · Question2.22 (b)FindtheprobabilitiesofeventsA;B;C;A\Bc; and A\CdeﬁnedinProblem2.2