Probability Lesson 2

www.MathAcademy.sg

Mr Ian Ang

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Venn diagram

Let Ω denote the entire probability space.

A ∪ B represents union =⇒ “Take everything in A and B”.A ∩ B represents intersection =⇒ “Take common parts in A and B”.

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Venn diagram

Let Ω denote the entire probability space.

A ∪ B represents union =⇒ “Take everything in A and B”.A ∩ B represents intersection =⇒ “Take common parts in A and B”.

c⃝ 2015 Math Academy www.MathAcademy.sg 3

Venn diagram

Let Ω denote the entire probability space.

A ∪ B represents union =⇒ “Take everything in A and B”.A ∩ B represents intersection =⇒ “Take common parts in A and B”.

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.Useful Results..

.

(i) P(A) + P(A′) = 1P(A|B) + P(A′|B) = 1.

Ω

A B

(ii) P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

Ω

A B

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.Useful Results..

.

(iii) P(A ∪ B) = P(A) + P(B ∩ A′)

Ω

A B

(iv) P(A) = P(A ∩ B) + P(A ∩ B ′)

Ω

A B

Recall also the following formula which would be useful for venn diagramquestions.

P(A|B) =P(A ∩ B)

P(B)or P(A|B) · P(B) = P(A ∩ B)

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.Useful Results..

.

(iii) P(A ∪ B) = P(A) + P(B ∩ A′)

Ω

A B

(iv) P(A) = P(A ∩ B) + P(A ∩ B ′)

Ω

A B

Recall also the following formula which would be useful for venn diagramquestions.

P(A|B) =P(A ∩ B)

P(B)or P(A|B) · P(B) = P(A ∩ B)

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Mutually Exclusive VS Independence

Two events, A and B, are said to be mutually exclusive if and only if the 2 setsare disjoint (no intersection between the 2 sets). That is,

P(A ∩ B) = 0.

Mutually exclusive events cannot happen at the same time.

Ω

A B

Mutually exclusive

Two events, A and B, are said to be independent, if theoccurence/non-occurence of A does not affect B. The concept of independencecannot be ‘seen’ on a venn-diagram. If A and B are independent, then

P(A ∩ B) = P(A) · P(B).

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Mutually Exclusive VS Independence

Two events, A and B, are said to be mutually exclusive if and only if the 2 setsare disjoint (no intersection between the 2 sets). That is,

P(A ∩ B) = 0.

Mutually exclusive events cannot happen at the same time.

Ω

A B

Mutually exclusive

Two events, A and B, are said to be independent, if theoccurence/non-occurence of A does not affect B. The concept of independencecannot be ‘seen’ on a venn-diagram. If A and B are independent, then

P(A ∩ B) = P(A) · P(B).

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Mutually Exclusive VS Independence

Two events, A and B, are said to be mutually exclusive if and only if the 2 setsare disjoint (no intersection between the 2 sets). That is,

P(A ∩ B) = 0.

Mutually exclusive events cannot happen at the same time.

Ω

A B

Mutually exclusive

Two events, A and B, are said to be independent, if theoccurence/non-occurence of A does not affect B. The concept of independencecannot be ‘seen’ on a venn-diagram. If A and B are independent, then

P(A ∩ B) = P(A) · P(B).

c⃝ 2015 Math Academy www.MathAcademy.sg 10

Mutually Exclusive VS Independence

Two events, A and B, are said to be mutually exclusive if and only if the 2 setsare disjoint (no intersection between the 2 sets). That is,

P(A ∩ B) = 0.

Mutually exclusive events cannot happen at the same time.

Ω

A B

Mutually exclusive

Two events, A and B, are said to be independent, if theoccurence/non-occurence of A does not affect B. The concept of independencecannot be ‘seen’ on a venn-diagram. If A and B are independent, then

P(A ∩ B) = P(A) · P(B).

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Mutually Exclusive VS Independence

Two events, A and B, are said to be mutually exclusive if and only if the 2 setsare disjoint (no intersection between the 2 sets). That is,

P(A ∩ B) = 0.

Mutually exclusive events cannot happen at the same time.

Ω

A B

Mutually exclusive

Two events, A and B, are said to be independent, if theoccurence/non-occurence of A does not affect B. The concept of independencecannot be ‘seen’ on a venn-diagram. If A and B are independent, then

P(A ∩ B) = P(A) · P(B).

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.What to test for?..

.

Mutually exclusive:

P(A ∩ B) = 0

Independent:

(1) P(A ∩ B) = P(A) · P(B)

(2) P(A|B) = P(A)

We can use either of the 2 equations to prove independence of two events.

c⃝ 2015 Math Academy www.MathAcademy.sg 13

.What to test for?..

.

Mutually exclusive:

P(A ∩ B) = 0

Independent:

(1) P(A ∩ B) = P(A) · P(B)

(2) P(A|B) = P(A)

We can use either of the 2 equations to prove independence of two events.

c⃝ 2015 Math Academy www.MathAcademy.sg 14

.What to test for?..

.

Mutually exclusive:

P(A ∩ B) = 0

Independent:

(1) P(A ∩ B) = P(A) · P(B)

(2) P(A|B) = P(A)

We can use either of the 2 equations to prove independence of two events.

c⃝ 2015 Math Academy www.MathAcademy.sg 15

.What to test for?..

.

Mutually exclusive:

P(A ∩ B) = 0

Independent:

(1) P(A ∩ B) = P(A) · P(B)

(2) P(A|B) = P(A)

We can use either of the 2 equations to prove independence of two events.

c⃝ 2015 Math Academy www.MathAcademy.sg 16

.What to test for?..

.

Mutually exclusive:

P(A ∩ B) = 0

Independent:

(1) P(A ∩ B) = P(A) · P(B)

(2) P(A|B) = P(A)

We can use either of the 2 equations to prove independence of two events.

c⃝ 2015 Math Academy www.MathAcademy.sg 17

.Example (6)..

.

A fair 6-sided die is thrown twice. Let A,B and C denote the following events:

A - The first throw is ‘1’.B - The first throw is ‘2’.C - The second throw is ‘1’.

Q. Are A and B mutually exclusive?A. Yes. The 2 events cannot happen at the same time (a throw cannot be both1 and 2). Hence P(A ∩ B) = 0.

Q. Are A and C independent?A. Yes. The outcome of the first throw does not affect the second throw.

Q. Are A and C mutually exclusive?A. No. A and C can both happen together. P(A ∩ C) = 1

6× 1

6= 1

36= 0.

c⃝ 2015 Math Academy www.MathAcademy.sg 18

.Example (6)..

.

A fair 6-sided die is thrown twice. Let A,B and C denote the following events:

A - The first throw is ‘1’.B - The first throw is ‘2’.C - The second throw is ‘1’.

Q. Are A and B mutually exclusive?A. Yes. The 2 events cannot happen at the same time (a throw cannot be both1 and 2). Hence P(A ∩ B) = 0.

Q. Are A and C independent?A. Yes. The outcome of the first throw does not affect the second throw.

Q. Are A and C mutually exclusive?A. No. A and C can both happen together. P(A ∩ C) = 1

6× 1

6= 1

36= 0.

c⃝ 2015 Math Academy www.MathAcademy.sg 19

.Example (6)..

.

A fair 6-sided die is thrown twice. Let A,B and C denote the following events:

A - The first throw is ‘1’.B - The first throw is ‘2’.C - The second throw is ‘1’.

Q. Are A and B mutually exclusive?A. Yes. The 2 events cannot happen at the same time (a throw cannot be both1 and 2). Hence P(A ∩ B) = 0.

Q. Are A and C independent?A. Yes. The outcome of the first throw does not affect the second throw.

Q. Are A and C mutually exclusive?A. No. A and C can both happen together. P(A ∩ C) = 1

6× 1

6= 1

36= 0.

c⃝ 2015 Math Academy www.MathAcademy.sg 20

.Example (6)..

.

A fair 6-sided die is thrown twice. Let A,B and C denote the following events:

A - The first throw is ‘1’.B - The first throw is ‘2’.C - The second throw is ‘1’.

Q. Are A and B mutually exclusive?A. Yes. The 2 events cannot happen at the same time (a throw cannot be both1 and 2). Hence P(A ∩ B) = 0.

Q. Are A and C independent?A. Yes. The outcome of the first throw does not affect the second throw.

Q. Are A and C mutually exclusive?A. No. A and C can both happen together. P(A ∩ C) = 1

6× 1

6= 1

36= 0.

c⃝ 2015 Math Academy www.MathAcademy.sg 21

.Example (7)..

.

For events A and B, it is given that P(A|B ′) = 23, P(A ∩ B) = 1

10and

P(B) = 25. Find,

(a) P(A ∩ B ′)

(b) P(A)

(c) State whether A and B are independent, giving a reason for your answer.

(a)

P(A|B ′) =P(A ∩ B ′)

P(B ′)

2

3=

P(A ∩ B ′)

1− P(B)

2

3=

P(A ∩ B ′)

1− 25

P(A ∩ B ′) =2

5

c⃝ 2015 Math Academy www.MathAcademy.sg 22

.Example (7)..

.

For events A and B, it is given that P(A|B ′) = 23, P(A ∩ B) = 1

10and

P(B) = 25. Find,

(a) P(A ∩ B ′)

(b) P(A)

(c) State whether A and B are independent, giving a reason for your answer.

(a)

P(A|B ′) =P(A ∩ B ′)

P(B ′)

2

3=

P(A ∩ B ′)

1− P(B)

2

3=

P(A ∩ B ′)

1− 25

P(A ∩ B ′) =2

5

c⃝ 2015 Math Academy www.MathAcademy.sg 23

.Example (7)..

.

For events A and B, it is given that P(A|B ′) = 23, P(A ∩ B) = 1

10and

P(B) = 25. Find,

(a) P(A ∩ B ′)

(b) P(A)

(c) State whether A and B are independent, giving a reason for your answer.

(b)

P(A) = P(A ∩ B) + P(A ∩ B ′)

=1

10+

2

5=

1

2

(c) P(A ∩ B) = 110

and P(A)·P(B) = 12· 25= 1

5.

Since P(A ∩ B) = P(A)·P(B), A and B are not independent.

c⃝ 2015 Math Academy www.MathAcademy.sg 24

.Example (7)..

.

For events A and B, it is given that P(A|B ′) = 23, P(A ∩ B) = 1

10and

P(B) = 25. Find,

(a) P(A ∩ B ′)

(b) P(A)

(c) State whether A and B are independent, giving a reason for your answer.

(b)

P(A) = P(A ∩ B) + P(A ∩ B ′)

=1

10+

2

5=

1

2

(c) P(A ∩ B) = 110

and P(A)·P(B) = 12· 25= 1

5.

Since P(A ∩ B) = P(A)·P(B), A and B are not independent.

c⃝ 2015 Math Academy www.MathAcademy.sg 25

.Example (7)..

.

For events A and B, it is given that P(A|B ′) = 23, P(A ∩ B) = 1

10and

P(B) = 25. Find,

(a) P(A ∩ B ′)

(b) P(A)

(c) State whether A and B are independent, giving a reason for your answer.

(b)

P(A) = P(A ∩ B) + P(A ∩ B ′)

=1

10+

2

5=

1

2

(c) P(A ∩ B) = 110

and P(A)·P(B) = 12· 25= 1

5.

Since P(A ∩ B) = P(A)·P(B), A and B are not independent.

c⃝ 2015 Math Academy www.MathAcademy.sg 26

.Example (8)..

.

Events A and B are such that P(A) = 17, P(B|A) = 1

5and P(A ∪ B) = 5

7. Find

P(A′ ∩ B ′) and P(B). State whether A and B are mutually exclusive, giving areason for your answer.

Ω

A B

P(A′)

Ω

A B

P(B ′)

Ω

A B

P(A′ ∩ B ′)

P(A′ ∩ B ′) = 1− P(A ∪ B)

= 1− 5

7

=2

7

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

5

7=

1

7+ P(B)− P(B|A) · P(A)

5

7=

1

7+ P(B)− 1

5· 17

P(B) =3

5

Since P(A ∩ B) = P(B|A) · P(A) = 15· 17= 0, A and B are not mutually

exclusive.

c⃝ 2015 Math Academy www.MathAcademy.sg 27

.Example (8)..

.

Events A and B are such that P(A) = 17, P(B|A) = 1

5and P(A ∪ B) = 5

7. Find

P(A′ ∩ B ′) and P(B). State whether A and B are mutually exclusive, giving areason for your answer.

Ω

A B

P(A′)

Ω

A B

P(B ′)

Ω

A B

P(A′ ∩ B ′)

P(A′ ∩ B ′) = 1− P(A ∪ B)

= 1− 5

7

=2

7

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

5

7=

1

7+ P(B)− P(B|A) · P(A)

5

7=

1

7+ P(B)− 1

5· 17

P(B) =3

5

Since P(A ∩ B) = P(B|A) · P(A) = 15· 17= 0, A and B are not mutually

exclusive.

c⃝ 2015 Math Academy www.MathAcademy.sg 28

.Example (8)..

.

Events A and B are such that P(A) = 17, P(B|A) = 1

5and P(A ∪ B) = 5

7. Find

P(A′ ∩ B ′) and P(B). State whether A and B are mutually exclusive, giving areason for your answer.

Ω

A B

P(A′)

Ω

A B

P(B ′)

Ω

A B

P(A′ ∩ B ′)

P(A′ ∩ B ′) = 1− P(A ∪ B)

= 1− 5

7

=2

7

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

5

7=

1

7+ P(B)− P(B|A) · P(A)

5

7=

1

7+ P(B)− 1

5· 17

P(B) =3

5

Since P(A ∩ B) = P(B|A) · P(A) = 15· 17= 0, A and B are not mutually

exclusive.

c⃝ 2015 Math Academy www.MathAcademy.sg 29

.Example (8)..

.

Events A and B are such that P(A) = 17, P(B|A) = 1

5and P(A ∪ B) = 5

7. Find

P(A′ ∩ B ′) and P(B). State whether A and B are mutually exclusive, giving areason for your answer.

Ω

A B

P(A′)

Ω

A B

P(B ′)

Ω

A B

P(A′ ∩ B ′)

P(A′ ∩ B ′) = 1− P(A ∪ B)

= 1− 5

7

=2

7

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

5

7=

1

7+ P(B)− P(B|A) · P(A)

5

7=

1

7+ P(B)− 1

5· 17

P(B) =3

5

Since P(A ∩ B) = P(B|A) · P(A) = 15· 17= 0, A and B are not mutually

exclusive.

c⃝ 2015 Math Academy www.MathAcademy.sg 30

.Example (9)..

.

Given that the two events A and B are such that P(A|B) = 23, P(A ∩ B ′) = 1

4

and P(A ∩ B) = 512,

(a) determine if A and B are independent,

(b) find P(A ∪ B).

[a) Yes b) 78]

(a)

P(A) = P(A ∩ B ′) + P(A ∩ B)

=1

4+

5

12

=2

3

Since P(A|B) = P(A) = 23, the two

events are independent.

(b)We first find P(B).

P(A ∩ B) = P(A) · P(B)

5

12=

2

3· P(B)

P(B) =5

8

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

=2

3+

5

8− 5

12

=7

8c⃝ 2015 Math Academy www.MathAcademy.sg 31

.Example (9)..

.

Given that the two events A and B are such that P(A|B) = 23, P(A ∩ B ′) = 1

4

and P(A ∩ B) = 512,

(a) determine if A and B are independent,

(b) find P(A ∪ B).

[a) Yes b) 78]

(a)

P(A) = P(A ∩ B ′) + P(A ∩ B)

=1

4+

5

12

=2

3

Since P(A|B) = P(A) = 23, the two

events are independent.

(b)We first find P(B).

P(A ∩ B) = P(A) · P(B)

5

12=

2

3· P(B)

P(B) =5

8

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

=2

3+

5

8− 5

12

=7

8c⃝ 2015 Math Academy www.MathAcademy.sg 32

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(a) i) Find the probability that three doctors are selected.ii) Suppose there are 4 positions in the group to be filled, namely, chair person,

vice-chair, secretary and executive. Find the probability that the chair,vice-chair and secretary are doctors.

(a) i)

P(3 doctors are selected) =20C3 ×15 C1

35C4=

855

2618

ii)

P( chair, vice-chair and secretary are doctors) =(20C3 × 3!)×15 C1

35C4 × 4!=

855

10472

c⃝ 2015 Math Academy www.MathAcademy.sg 33

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(a) i) Find the probability that three doctors are selected.ii) Suppose there are 4 positions in the group to be filled, namely, chair person,

vice-chair, secretary and executive. Find the probability that the chair,vice-chair and secretary are doctors.

(a) i)

P(3 doctors are selected) =20C3 ×15 C1

35C4=

855

2618

ii)

P( chair, vice-chair and secretary are doctors) =(20C3 × 3!)×15 C1

35C4 × 4!=

855

10472

c⃝ 2015 Math Academy www.MathAcademy.sg 34

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(a) i) Find the probability that three doctors are selected.ii) Suppose there are 4 positions in the group to be filled, namely, chair person,

vice-chair, secretary and executive. Find the probability that the chair,vice-chair and secretary are doctors.

(a) i)

P(3 doctors are selected) =20C3 ×15 C1

35C4=

855

2618

ii)

P( chair, vice-chair and secretary are doctors) =(20C3 × 3!)×15 C1

35C4 × 4!=

855

10472

c⃝ 2015 Math Academy www.MathAcademy.sg 35

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(a) i) Find the probability that three doctors are selected.ii) Suppose there are 4 positions in the group to be filled, namely, chair person,

vice-chair, secretary and executive. Find the probability that the chair,vice-chair and secretary are doctors.

(a) i)

P(3 doctors are selected) =20C3 ×15 C1

35C4=

855

2618

ii)

P( chair, vice-chair and secretary are doctors) =(20C3 × 3!)×15 C1

35C4 × 4!=

855

10472

c⃝ 2015 Math Academy www.MathAcademy.sg 36

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(b) Given that two women are selected, find the probability that both of themare doctors.

(b)

P(Both are doctors | 2 women selected)

=P(Both women selected are doctors)

P(2 women selected)

=12C2 ×18 C2

35C4÷

17C2 ×18 C2

35C4

=33

68

c⃝ 2015 Math Academy www.MathAcademy.sg 37

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(b) Given that two women are selected, find the probability that both of themare doctors.

(b)

P(Both are doctors | 2 women selected)

=P(Both women selected are doctors)

P(2 women selected)

=12C2 ×18 C2

35C4÷

17C2 ×18 C2

35C4

=33

68

c⃝ 2015 Math Academy www.MathAcademy.sg 38

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(b) Given that two women are selected, find the probability that both of themare doctors.

(b)

P(Both are doctors | 2 women selected)

=P(Both women selected are doctors)

P(2 women selected)

=12C2 ×18 C2

35C4÷

17C2 ×18 C2

35C4

=33

68

c⃝ 2015 Math Academy www.MathAcademy.sg 39

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(b) Given that two women are selected, find the probability that both of themare doctors.

(b)

P(Both are doctors | 2 women selected)

=P(Both women selected are doctors)

P(2 women selected)

=12C2 ×18 C2

35C4÷

17C2 ×18 C2

35C4

=33

68

c⃝ 2015 Math Academy www.MathAcademy.sg 40

.Example (10)..

.

There are 20 doctors and 15 engineers attending a conference. The number ofwomen doctors and that of women engineers are 12 and 5 respectively. Fourparticipants from this group are selected randomly to chair some sessions ofpanel discussions.

(b) Given that two women are selected, find the probability that both of themare doctors.

(b)

P(Both are doctors | 2 women selected)

=P(Both women selected are doctors)

P(2 women selected)

=12C2 ×18 C2

35C4÷

17C2 ×18 C2

35C4

=33

68

c⃝ 2015 Math Academy www.MathAcademy.sg 41