1. Binomial Probability

2. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;

3. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B

4. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q

5. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB

6. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB

7. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB
 2
p
 pq
 pq
 2
q

8. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB
 2
p
 pq
 pq
 2
q
AAA
AAB
ABA
ABB
BAA
BAB
BBA
BBB

9. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB
 2
p
 pq
 pq
 2
q
AAA
AAB
ABA
ABB
BAA
BAB
BBA
BBB
 3
p
 qp2
 qp2
 2
pq
 qp2
 2
pq
 2
pq
 3
p

10. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB
 2
p
 pq
 pq
 2
q
AAA
AAB
ABA
ABB
BAA
BAB
BBA
BBB
 3
p
 qp2
 qp2
 2
pq
 qp2
 2
pq
 2
pq
 3
p
AAAA
AAAB
AABA
AABB
ABAA
ABAB
ABBA
ABBB
BAAA
BAAB
BABA
BABB
BBAA
BBAB
BBBA
BBBB

11. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then the
probability distribution is as follows;
A
B
 p
 q
AA
AB
BA
BB
 2
p
 pq
 pq
 2
q
AAA
AAB
ABA
ABB
BAA
BAB
BBA
BBB
 3
p
 qp2
 qp2
 2
pq
 qp2
 2
pq
 2
pq
 3
p
AAAA
AAAB
AABA
AABB
ABAA
ABAB
ABBA
ABBB
BAAA
BAAB
BABA
BABB
BBAA
BBAB
BBBA
BBBB
 4
p
 qp3
 qp3
 22
qp
 qp3
 22
qp
 22
qp
 3
pq
 qp3
 22
qp
 22
qp
 3
pq
 22
qp
 3
pq
 3
pq 4
q

12. 1 Event

13. 1 Event
  pAP 
  qBP 

14. 1 Event
  pAP 
  qBP 
2 Events

15. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 

16. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events

17. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events
  3
pAAAP 
  qpAandBP 2
32 
  2
32 pqBAandP 
  3
qBBBP 

18. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events
  3
pAAAP 
  qpAandBP 2
32 
  2
32 pqBAandP 
  3
qBBBP 
4 Events

19. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events
  3
pAAAP 
  qpAandBP 2
32 
  2
32 pqBAandP 
  3
qBBBP 
4 Events
  4
pAAAAP 
  qpAandBP 3
43 
  22
622 qpBAandP 
  3
43 pqBAandP 
  4
qBBBBP 

20. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events
  3
pAAAP 
  qpAandBP 2
32 
  2
32 pqBAandP 
  3
qBBBP 
4 Events
  4
pAAAAP 
  qpAandBP 3
43 
  22
622 qpBAandP 
  3
43 pqBAandP 
  4
qBBBBP 
 
 
is;timesexactlyoccur
willy thatprobabilitthen theand
andtimesrepeatediseventanIf
k
XqXP
pXPn



21. 1 Event
  pAP 
  qBP 
2 Events
  2
pAAP 
  pqAandBP 2
  2
qBBP 
3 Events
  3
pAAAP 
  qpAandBP 2
32 
  2
32 pqBAandP 
  3
qBBBP 
4 Events
  4
pAAAAP 
  qpAandBP 3
43 
  22
622 qpBAandP 
  3
43 pqBAandP 
  4
qBBBBP 
 
 
is;timesexactlyoccur
willy thatprobabilitthen theand
andtimesrepeatediseventanIf
k
XqXP
pXPn


  kkn
k
n
pqCkXP 

Note: X = k, means X will occur exactly k times

22. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?

23. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP

24. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
Let X be the number of black balls drawn

25. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

Let X be the number of black balls drawn

26. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

Let X be the number of black balls drawn

27. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
Let X be the number of black balls drawn

28. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
Let X be the number of black balls drawn

29. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

Let X be the number of black balls drawn

30. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

Let X be the number of black balls drawn

31. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?
Let X be the number of black balls drawn

32. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?
Let X be the number of black balls drawn
Let X be the number
favouring Party A

33. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?  
32
3
5
10
3
10
7
3 










 CAP
Let X be the number of black balls drawn
Let X be the number
favouring Party A

34. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?  
32
3
5
10
3
10
7
3 










 CAP
5
32
3
5
10
37C

Let X be the number of black balls drawn
Let X be the number
favouring Party A

35. e.g.(i) A bag contains 30 black balls and 20 white balls.
Seven drawings are made (with replacement), what is the
probability of drawing;
a) All black balls?
 
70
7
7
5
3
5
2
7 










 CXP
7
7
7
7
5
3C

78125
2187

b) 4 black balls?
 
43
4
7
5
3
5
2
4 










 CXP
7
43
4
7
5
32C

15625
4536

(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that;
a) 3 favoured Party A?  
32
3
5
10
3
10
7
3 










 CAP
5
32
3
5
10
37C

10000
1323

Let X be the number of black balls drawn
Let X be the number
favouring Party A

36. b) majority favour A?

37. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP

38. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 


39. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 

25000
4077


40. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 

25000
4077

c) at most 2 favoured A?

41. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 

25000
4077

c) at most 2 favoured A?
   312  XPXP

42. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 

25000
4077

c) at most 2 favoured A?
   312  XPXP
25000
4077
1

43. b) majority favour A?
 
50
5
5
41
4
5
32
3
5
10
3
10
7
10
3
10
7
10
3
10
7
3 
































 CCCXP
5
5
5
54
4
532
3
5
10
33737 CCC 

25000
4077

c) at most 2 favoured A?
   312  XPXP
25000
4077
1
25000
20923


44. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is
3
2
(i) Show that the probability that Megan earns one point for a
given weekend is , correct to four decimal places.79010

45. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is
3
2
(i) Show that the probability that Megan earns one point for a
given weekend is , correct to four decimal places.79010
Let X be the number of matches picked correctly

46. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is
3
2
(i) Show that the probability that Megan earns one point for a
given weekend is , correct to four decimal places.79010
Let X be the number of matches picked correctly
 
50
5
5
41
4
5
32
3
5
3
2
3
1
3
2
3
1
3
2
3
1
3 
































 CCCXP

47. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is
3
2
(i) Show that the probability that Megan earns one point for a
given weekend is , correct to four decimal places.79010
Let X be the number of matches picked correctly
 
50
5
5
41
4
5
32
3
5
3
2
3
1
3
2
3
1
3
2
3
1
3 
































 CCCXP
5
5
5
54
4
53
3
5
3
222 CCC 


48. 2005 Extension 1 HSC Q6a)
There are five matches on each weekend of a football season.
Megan takes part in a competition in which she earns 1 point if
she picks more than half of the winning teams for a weekend, and
zero points otherwise.
The probability that Megan correctly picks the team that wins in
any given match is
3
2
(i) Show that the probability that Megan earns one point for a
given weekend is , correct to four decimal places.79010
Let X be the number of matches picked correctly
 
50
5
5
41
4
5
32
3
5
3
2
3
1
3
2
3
1
3
2
3
1
3 
































 CCCXP
5
5
5
54
4
53
3
5
3
222 CCC 

79010

49. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.

50. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point

51. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP

52. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP
dp)2(to010

53. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP
dp)2(to010
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.

54. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP
dp)2(to010
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
   17116  YPYP

55. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP
dp)2(to010
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
   17116  YPYP
       180
18
18171
17
18
790102099079010209901  CC

56. (ii) Hence find the probability that Megan earns one point every
week of the eighteen week season. Give your answer correct
to two decimal places.
Let Y be the number of weeks Megan earns a point
     180
18
18
790102099018  CYP
dp)2(to010
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
   17116  YPYP
dp)2(to920
       180
18
18171
17
18
790102099079010209901  CC

57. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?

58. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
 2 green 0.1 0.1
0.01
P  


59. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
 2 green 0.1 0.1
0.01
P  

(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.

60. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
Let X be the number of people with green eyes
 2 green 0.1 0.1
0.01
P  

(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.

61. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
Let X be the number of people with green eyes
     
18 220
22 0.9 0.1P X C 
 2 green 0.1 0.1
0.01
P  

(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.

62. 2007 Extension 1 HSC Q4a)
In a large city, 10% of the population has green eyes.
(i) What is the probability that two randomly chosen people have
green eyes?
Let X be the number of people with green eyes
     
18 220
22 0.9 0.1P X C 
0.2851
0.285 (to 3 dp)


 2 green 0.1 0.1
0.01
P  

(ii) What is the probability that exactly two of a group of 20 randomly
chosen people have green eyes? Give your answer correct to three
decimal eyes.

63. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.

64. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
   2 1 2P X P X   

65. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
   2 1 2P X P X   
           
18 2 19 1 20 020 20 20
2 1 01 0 9 0 1 0 9 0 1 0 9 0 1C C C         

66. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
   2 1 2P X P X   
0.3230
0 32 (to 2 dp)

 
           
18 2 19 1 20 020 20 20
2 1 01 0 9 0 1 0 9 0 1 0 9 0 1C C C         

67. (iii) What is the probability that more than two of a group of 20
randomly chosen people have green eyes? Give your answer
correct to two decimal places.
   2 1 2P X P X   
0.3230
0 32 (to 2 dp)

 
           
18 2 19 1 20 020 20 20
2 1 01 0 9 0 1 0 9 0 1 0 9 0 1C C C         
Exercise 10J;
1 to 24 even