Slides for Probability lecture 2. JC H2 Maths A levels Singapore www.mathacademy.sg

1. Probability Lesson 2
www.MathAcademy.sg
Mr Ian Ang
c⃝ 2015 Math Academy www.MathAcademy.sg 1

2. 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 2

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”.
c⃝ 2015 Math Academy www.MathAcademy.sg 3

4. 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 4

5. .
Useful Results
..
.
(i) P(A) + P(A′
) = 1
P(A|B) + P(A′
|B) = 1.
Ω
A B
(ii) P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Ω
A B
c⃝ 2015 Math Academy www.MathAcademy.sg 5

6. .
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 diagram
questions.
P(A|B) =
P(A ∩ B)
P(B)
or P(A|B) · P(B) = P(A ∩ B)
c⃝ 2015 Math Academy www.MathAcademy.sg 6

7. .
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 diagram
questions.
P(A|B) =
P(A ∩ B)
P(B)
or P(A|B) · P(B) = P(A ∩ B)
c⃝ 2015 Math Academy www.MathAcademy.sg 7

8. Mutually Exclusive VS Independence
Two events, A and B, are said to be mutually exclusive if and only if the 2 sets
are 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 the
occurence/non-occurence of A does not affect B. The concept of independence
cannot 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 8

9. Mutually Exclusive VS Independence
Two events, A and B, are said to be mutually exclusive if and only if the 2 sets
are 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 the
occurence/non-occurence of A does not affect B. The concept of independence
cannot 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 9

10. Mutually Exclusive VS Independence
Two events, A and B, are said to be mutually exclusive if and only if the 2 sets
are 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 the
occurence/non-occurence of A does not affect B. The concept of independence
cannot 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

11. Mutually Exclusive VS Independence
Two events, A and B, are said to be mutually exclusive if and only if the 2 sets
are 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 the
occurence/non-occurence of A does not affect B. The concept of independence
cannot 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 11

12. Mutually Exclusive VS Independence
Two events, A and B, are said to be mutually exclusive if and only if the 2 sets
are 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 the
occurence/non-occurence of A does not affect B. The concept of independence
cannot 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 12

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 13

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 14

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 15

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 16

17. .
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

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 both
1 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

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 both
1 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

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 both
1 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

21. .
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 both
1 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

22. .
Example (7)
..
.
For events A and B, it is given that P(A|B′
) = 2
3
, P(A ∩ B) = 1
10
and
P(B) = 2
5
. 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 − 2
5
P(A ∩ B′
) =
2
5
c⃝ 2015 Math Academy www.MathAcademy.sg 22

23. .
Example (7)
..
.
For events A and B, it is given that P(A|B′
) = 2
3
, P(A ∩ B) = 1
10
and
P(B) = 2
5
. 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 − 2
5
P(A ∩ B′
) =
2
5
c⃝ 2015 Math Academy www.MathAcademy.sg 23

24. .
Example (7)
..
.
For events A and B, it is given that P(A|B′
) = 2
3
, P(A ∩ B) = 1
10
and
P(B) = 2
5
. 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) = 1
10
and P(A)·P(B) = 1
2
· 2
5
= 1
5
.
Since P(A ∩ B) ̸= P(A)·P(B), A and B are not independent.
c⃝ 2015 Math Academy www.MathAcademy.sg 24

25. .
Example (7)
..
.
For events A and B, it is given that P(A|B′
) = 2
3
, P(A ∩ B) = 1
10
and
P(B) = 2
5
. 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) = 1
10
and P(A)·P(B) = 1
2
· 2
5
= 1
5
.
Since P(A ∩ B) ̸= P(A)·P(B), A and B are not independent.
c⃝ 2015 Math Academy www.MathAcademy.sg 25

26. .
Example (7)
..
.
For events A and B, it is given that P(A|B′
) = 2
3
, P(A ∩ B) = 1
10
and
P(B) = 2
5
. 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) = 1
10
and P(A)·P(B) = 1
2
· 2
5
= 1
5
.
Since P(A ∩ B) ̸= P(A)·P(B), A and B are not independent.
c⃝ 2015 Math Academy www.MathAcademy.sg 26

27. .
Example (8)
..
.
Events A and B are such that P(A) = 1
7
, P(B|A) = 1
5
and P(A ∪ B) = 5
7
. Find
P(A′
∩ B′
) and P(B). State whether A and B are mutually exclusive, giving a
reason 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
·
1
7
P(B) =
3
5
Since P(A ∩ B) = P(B|A) · P(A) = 1
5
· 1
7
̸= 0, A and B are not mutually
exclusive.
c⃝ 2015 Math Academy www.MathAcademy.sg 27

28. .
Example (8)
..
.
Events A and B are such that P(A) = 1
7
, P(B|A) = 1
5
and P(A ∪ B) = 5
7
. Find
P(A′
∩ B′
) and P(B). State whether A and B are mutually exclusive, giving a
reason 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
·
1
7
P(B) =
3
5
Since P(A ∩ B) = P(B|A) · P(A) = 1
5
· 1
7
̸= 0, A and B are not mutually
exclusive.
c⃝ 2015 Math Academy www.MathAcademy.sg 28

29. .
Example (8)
..
.
Events A and B are such that P(A) = 1
7
, P(B|A) = 1
5
and P(A ∪ B) = 5
7
. Find
P(A′
∩ B′
) and P(B). State whether A and B are mutually exclusive, giving a
reason 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
·
1
7
P(B) =
3
5
Since P(A ∩ B) = P(B|A) · P(A) = 1
5
· 1
7
̸= 0, A and B are not mutually
exclusive.
c⃝ 2015 Math Academy www.MathAcademy.sg 29

30. .
Example (8)
..
.
Events A and B are such that P(A) = 1
7
, P(B|A) = 1
5
and P(A ∪ B) = 5
7
. Find
P(A′
∩ B′
) and P(B). State whether A and B are mutually exclusive, giving a
reason 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
·
1
7
P(B) =
3
5
Since P(A ∩ B) = P(B|A) · P(A) = 1
5
· 1
7
̸= 0, A and B are not mutually
exclusive.
c⃝ 2015 Math Academy www.MathAcademy.sg 30

31. .
Example (9)
..
.
Given that the two events A and B are such that P(A|B) = 2
3
, P(A ∩ B′
) = 1
4
and P(A ∩ B) = 5
12
,
(a) determine if A and B are independent,
(b) find P(A ∪ B).
[a) Yes b) 7
8
]
(a)
P(A) = P(A ∩ B′
) + P(A ∩ B)
=
1
4
+
5
12
=
2
3
Since P(A|B) = P(A) = 2
3
, 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
8
c⃝ 2015 Math Academy www.MathAcademy.sg 31

32. .
Example (9)
..
.
Given that the two events A and B are such that P(A|B) = 2
3
, P(A ∩ B′
) = 1
4
and P(A ∩ B) = 5
12
,
(a) determine if A and B are independent,
(b) find P(A ∪ B).
[a) Yes b) 7
8
]
(a)
P(A) = P(A ∩ B′
) + P(A ∩ B)
=
1
4
+
5
12
=
2
3
Since P(A|B) = P(A) = 2
3
, 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
8
c⃝ 2015 Math Academy www.MathAcademy.sg 32

33. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel 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

34. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel 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

35. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel 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

36. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel 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

37. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel discussions.
(b) Given that two women are selected, find the probability that both of them
are doctors.
(b)
P(Both are doctors | 2 women selected)
=
P(Both women selected are doctors)
P(2 women selected)
=
12
C2 ×18
C2
35C4
÷
17
C2 ×18
C2
35C4
=
33
68
c⃝ 2015 Math Academy www.MathAcademy.sg 37

38. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel discussions.
(b) Given that two women are selected, find the probability that both of them
are doctors.
(b)
P(Both are doctors | 2 women selected)
=
P(Both women selected are doctors)
P(2 women selected)
=
12
C2 ×18
C2
35C4
÷
17
C2 ×18
C2
35C4
=
33
68
c⃝ 2015 Math Academy www.MathAcademy.sg 38

39. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel discussions.
(b) Given that two women are selected, find the probability that both of them
are doctors.
(b)
P(Both are doctors | 2 women selected)
=
P(Both women selected are doctors)
P(2 women selected)
=
12
C2 ×18
C2
35C4
÷
17
C2 ×18
C2
35C4
=
33
68
c⃝ 2015 Math Academy www.MathAcademy.sg 39

40. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel discussions.
(b) Given that two women are selected, find the probability that both of them
are doctors.
(b)
P(Both are doctors | 2 women selected)
=
P(Both women selected are doctors)
P(2 women selected)
=
12
C2 ×18
C2
35C4
÷
17
C2 ×18
C2
35C4
=
33
68
c⃝ 2015 Math Academy www.MathAcademy.sg 40

41. .
Example (10)
..
.
There are 20 doctors and 15 engineers attending a conference. The number of
women doctors and that of women engineers are 12 and 5 respectively. Four
participants from this group are selected randomly to chair some sessions of
panel discussions.
(b) Given that two women are selected, find the probability that both of them
are doctors.
(b)
P(Both are doctors | 2 women selected)
=
P(Both women selected are doctors)
P(2 women selected)
=
12
C2 ×18
C2
35C4
÷
17
C2 ×18
C2
35C4
=
33
68
c⃝ 2015 Math Academy www.MathAcademy.sg 41