The document discusses binomial probability distributions. It explains that if an event has two possible outcomes and is repeated, the probability of each outcome follows a binomial distribution. It provides examples of calculating binomial probabilities for 1, 2, 3, and 4 events. The key points are: - Binomial probabilities use the formula P(X=k) = nCk * pk * (1-p)(n-k) - Where X is the number of successes, n is the number of trials, p is the probability of success on each trial, and q is the probability of failure. - It illustrates calculating binomial probabilities for examples like drawing balls from a bag.

1. Binomial Probability

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

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

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

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

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

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

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

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

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

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

12. 1 Event

13. 1 Event
P  A  p
P B   q

14. 1 Event 2 Events
P  A  p
P B   q

15. 1 Event 2 Events
P  A  p P AA  p 2
P B   q P AandB   2 pq
PBB   q 2

16. 1 Event 2 Events 3 Events
P  A  p P AA  p 2
P B   q P AandB   2 pq
PBB   q 2

17. 1 Event 2 Events 3 Events
P  A  p P AA  p 2 P AAA   p 3
P B   q P AandB   2 pq P2 AandB   3 p 2 q
PBB   q 2 P Aand 2 B   3 pq 2
PBBB  q 3

18. 1 Event 2 Events 3 Events
P  A  p P AA  p 2 P AAA   p 3
P B   q P AandB   2 pq P2 AandB   3 p 2 q
PBB   q 2 P Aand 2 B   3 pq 2
PBBB  q 3
4 Events

19. 1 Event 2 Events 3 Events
P  A  p P AA  p 2 P AAA   p 3
P B   q P AandB   2 pq P2 AandB   3 p 2 q
PBB   q 2 P Aand 2 B   3 pq 2
PBBB  q 3
4 Events
P AAAA   p 4
P3 AandB   4 p 3q
P2 Aand 2 B   6 p 2 q 2
P Aand 3B   4 pq 3
PBBBB  q 4

20. 1 Event 2 Events 3 Events
P  A  p P AA  p 2 P AAA   p 3
P B   q P AandB   2 pq P2 AandB   3 p 2 q
PBB   q 2 P Aand 2 B   3 pq 2
PBBB  q 3
4 Events If an event is repeated n times and P X   p
P AAAA   p 4 and P X   q then the probabilit y that X will
P3 AandB   4 p 3q occur exactly k times is;
P2 Aand 2 B   6 p 2 q 2
P Aand 3B   4 pq 3
PBBBB  q 4

21. 1 Event 2 Events 3 Events
P  A  p P AA  p 2 P AAA   p 3
P B   q P AandB   2 pq P2 AandB   3 p 2 q
PBB   q 2 P Aand 2 B   3 pq 2
PBBB  q 3
4 Events If an event is repeated n times and P X   p
P AAAA   p 4 and P X   q then the probabilit y that X will
P3 AandB   4 p 3q occur exactly k times is;
P2 Aand 2 B   6 p 2 q 2
P X  k  nCk q nk p k
P Aand 3B   4 pq 3
PBBBB  q 4 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?
0 7
P X  7  7C7    
2 3
   
 5 5

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; Let X be the number of black balls drawn
a) All black balls?
0 7
P X  7  7C7    
2  3

 5 5

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; Let X be the number of black balls drawn
a) All black balls?
0 7
P X  7  7C7    
2  3

 5 5
7
C7 37
 7
5

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; Let X be the number of black balls drawn
a) All black balls?
0 7
P X  7  7C7    
2  3

 5 5
7
C7 37
 7
5
2187

78125

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7
P X  7  7C7    
2  3

 5 5
7
C7 37
 7
5
2187

78125

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37
 7
5
2187

78125

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187

78125

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625
(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. 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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625
(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
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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A5C3    
7 3
a) 3 favoured Party A?    
 10   10 
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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A5C3    
7 3
a) 3 favoured Party A?    
 10   10 
Let X be the number 5
C3 7 233
favouring Party A 
105

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; Let X be the number of black balls drawn
a) All black balls? b) 4 black balls?
0 7 3 4
P X  7  7C7    
2  3
P X  4 7C4    
2  3
 
 5 5  5 5
7
C7 37 7
C4 2334
 7 
5 57
2187 4536
 
78125 15625
(ii) At an election 30% of voters favoured Party A.
If at random an interviewer selects 5 voters, what is the
probability that; 2 3
P3 A5C3    
7 3
a) 3 favoured Party A?    
 10   10 
Let X be the number 5
C3 7 233 1323
favouring Party A  5

10 10000

36. b) majority favour A?

37. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 

38. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105

39. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105
4077

25000

40. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105
4077

25000
c) at most 2 favoured A?

41. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105
4077

25000
c) at most 2 favoured A?
P X  2  1  P X  3

42. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105
4077

25000
c) at most 2 favoured A?
P X  2  1  P X  3
4077
 1
25000

43. b) majority favour A?
2 3 1 4 0 5
P X  35C3      5C4      5C5    
7 3 7 3 7 3
          
 10   10   10   10   10   10 
5
C3 7 233  5C4 7  34  5C5 35

105
4077

25000
c) at most 2 favoured A?
P X  2  1  P X  3
4077
 1
25000
20923

25000

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 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.

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 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
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 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X  35C3      5C4      5C5    
1 2 1 2 1 2
          
 3  3   3  3   3  3 

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 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X  35C3      5C4      5C5    
1 2 1 2 1 2
          
 3  3   3  3   3  3 
5
C3 23  5C4 2 4  5C5 25

35

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 2
3
(i) Show that the probability that Megan earns one point for a
given weekend is 0 7901, correct to four decimal places.
Let X be the number of matches picked correctly
2 3 1 4 0 5
P X  35C3      5C4      5C5    
1 2 1 2 1 2
          
 3  3   3  3   3  3 
5
C3 23  5C4 2 4  5C5 25

35
 0 7901

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
PY  1818C18 0  2099 0  7901
0 18

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
PY  1818C18 0  2099 0  7901
0 18
 0 01 (to 2 dp)

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
PY  1818C18 0  2099 0  7901
0 18
 0 01 (to 2 dp)
(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
PY  1818C18 0  2099 0  7901
0 18
 0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY  16  1  PY  17 

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
PY  1818C18 0  2099 0  7901
0 18
 0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY  16  1  PY  17 
 118C17 0  2099 0  7901 18C18 0  2099 0  7901
1 17 0 18

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
PY  1818C18 0  2099 0  7901
0 18
 0 01 (to 2 dp)
(iii) Find the probability that Megan earns at most 16 points
during the eighteen week season. Give your answer correct
to two decimal places.
PY  16  1  PY  17 
 118C17 0  2099 0  7901 18C18 0  2099 0  7901
1 17 0 18
 0 92 (to 2 dp)

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?
P  2 green   0.1 0.1
 0.01

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?
P  2 green   0.1 0.1
 0.01
(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?
P  2 green   0.1 0.1
 0.01
(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.
Let X be the number of people with green 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?
P  2 green   0.1 0.1
 0.01
(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.
Let X be the number of people with green eyes
P  X  2   C2  0.9   0.1
20 18 2

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?
P  2 green   0.1 0.1
 0.01
(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.
Let X be the number of people with green eyes
P  X  2   C2  0.9   0.1
20 18 2
 0.2851
 0.285 (to 3 dp)

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.
P  X  2  1  P  X  2

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.
P  X  2  1  P  X  2
 1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1
20 18 2 20 19 1 20 20 0

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.
P  X  2  1  P  X  2
 1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1
20 18 2 20 19 1 20 20 0
 0.3230
 0  32 (to 2 dp)

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.
P  X  2  1  P  X  2
 1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1
20 18 2 20 19 1 20 20 0
 0.3230
 0  32 (to 2 dp)
Exercise 10J;
1 to 24 even