This document discusses methods for counting outcomes and possibilities, including tree diagrams and the Fundamental Counting Principle. It provides examples of using each method to determine the number of possible uniforms for a football team and lunch specials at a deli. The document also introduces factorials as a way to write the calculation for arrangement possibilities more concisely.

1. QA06 PROBABILITY
9/11/2013 1

2. 9/11/2013 2

3. 9/11/2013 3

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5. 9/11/2013 5

6. 1) tree diagram
2) sample space
3) event
4) Fundamental Counting Principle
5) Factorial
Counting Outcomes
 Count outcomes using a tree diagram.
 Count outcomes using the
Fundamental Counting Principle.

7. Counting Outcomes
One method used for counting the number of possible outcomes is to draw a
tree diagram.
The last column of a tree diagram shows all of the possible outcomes.
The list of all possible outcomes is called the sample space, while any collection
of one or more outcomes in the sample space is called an event.

8. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.

9. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.
Jersey Pants Shoes Outcomes
Red
White
Gray

10. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.
Jersey Pants Shoes Outcomes
Red
White
Gray
Gray
Black
Gray
Black
Gray
Black

11. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.
Jersey Pants Shoes Outcomes
Red
White
Gray
Gray
Black
Gray
Black
Gray
Black
Black
White
Black
White
Black
White
Black
White
Black
White
Black
White

12. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.
Jersey Pants Shoes Outcomes
Red
White
Gray
Gray
Black
Gray
Black
Gray
Black
Black
White
Black
White
RGB
RGW
RBB
RBW
WGB
WGW
WBB
WBW
GGB
GGW
GBB
GBW
Black
White
Black
White
Black
White
Black
White

13. Counting Outcomes
A football team uses red jerseys for road games, white jerseys for home games,
and gray jerseys for practice games. The team uses gray or black pants, and
black and white shoes.
Use a tree diagram to determine the number of possible uniforms.
Jersey Pants Shoes Outcomes
Red
White
Gray
Gray
Black
Gray
Black
Gray
Black
Black
White
Black
White
RGB
RGW
RBB
RBW
WGB
WGW
WBB
WBW
GGB
GGW
GBB
GBW
Black
White
Black
White
Black
White
Black
White
The tree diagram shows that there are 12 possible outcomes.

14. Counting Outcomes
In the previous example, the number of possible outcomes could also be found by
multiplying the number of choices for each item.

15. Counting Outcomes
In the previous example, the number of possible outcomes could also be found by
multiplying the number of choices for each item.
The team could choose from:
3
different
colored jerseys
2
different
colored pants
2
different
colored shoes

16. Counting Outcomes
In the previous example, the number of possible outcomes could also be found by
multiplying the number of choices for each item.
The team could choose from:
3
different
colored jerseys
2
different
colored pants
2
different
colored shoes
There are 3 X 2 X 2 or 12 possible uniforms.

17. Counting Outcomes
In the previous example, the number of possible outcomes could also be found by
multiplying the number of choices for each item.
The team could choose from:
3
different
colored jerseys
2
different
colored pants
2
different
colored shoes
There are 3 X 2 X 2 or 12 possible uniforms.
This example illustrates the Fundamental Counting Principle.

18. Counting Outcomes
In the previous example, the number of possible outcomes could also be found by
multiplying the number of choices for each item.
The team could choose from:
3
different
colored jerseys
2
different
colored pants
2
different
colored shoes
There are 3 X 2 X 2 or 12 possible uniforms.
This example illustrates the Fundamental Counting Principle.
If an event M can occur in m ways, and is followed by an event N that can occur
in n ways, then the event M followed by event N can occur m X n ways.

19. Counting Outcomes
A deli offers a lunch special in which you can choose a sandwich, a side dish, an a
beverage.
If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages,
from which to choose, how many different lunch specials can be ordered?

20. Counting Outcomes
A deli offers a lunch special in which you can choose a sandwich, a side dish, an a
beverage.
If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages,
from which to choose, how many different lunch specials can be ordered?
Multiply to find the number of lunch specials.
sandwich
choices
side dish
choices
beverage
choices
number of
specialsX X =

21. Counting Outcomes
A deli offers a lunch special in which you can choose a sandwich, a side dish, an a
beverage.
If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages,
from which to choose, how many different lunch specials can be ordered?
Multiply to find the number of lunch specials.
sandwich
choices
side dish
choices
beverage
choices
number of
specialsX X =
10 X 12 X 7 = 840
The number of different lunch specials is 840.

22. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?

23. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.

24. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.

25. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.
 After choosing a game for the first position, there are nine games left from which
to choose for the second position.

26. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.
 After choosing a game for the first position, there are nine games left from which
to choose for the second position.
 There are now eight choices for the third position.

27. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.
 After choosing a game for the first position, there are nine games left from which
to choose for the second position.
 There are now eight choices for the third position.
 This process continues until there is only one choice left for the last position.

28. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.
 After choosing a game for the first position, there are nine games left from which
to choose for the second position.
 There are now eight choices for the third position.
 This process continues until there is only one choice left for the last position.
Let n represent the number of arrangements.
n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800

29. Counting Outcomes
A.J. is setting up a display of the ten most popular video games from the previous
week. If he places the games side-by-side on a shelf, in how many different ways
can he arrange them?
The number of ways to arrange the games can be found by multiplying the number
of choices for each position.
 A.J. has 10 games from which to choose for the first position.
 After choosing a game for the first position, there are nine games left from which
to choose for the second position.
 There are now eight choices for the third position.
 This process continues until there is only one choice left for the last position.
Let n represent the number of arrangements.
n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800
There are 3,628,800 different ways to arrange the video games.

30. Counting Outcomes
The expression n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 used in the previous example
can be written as 10! using a factorial.
The expression n!, read n factorial, where n is greater than zero, is the
product of all positive integers beginning with n and counting backward to 1.
n! = n(n – 1)* (n – 2) * . . . 3 * 2 * 1
Example: 5! = 5 * 4 * 3 * 2 * 1 or 120

31. Counting Outcomes
0! is defined as being equal to 1. Let’s see why.

32. Counting Outcomes
0! is defined as being equal to 1. Let’s see why.
4
!4
!3
Writing this out using the definition of factorials.
4
1*2*3*4
4
1*2*3*4
!3

33. Counting Outcomes
0! is defined as being equal to 1. Let’s see why.
4
!4
!3
Writing this out using the definition of factorials.
4
1*2*3*4
4
1*2*3*4
!3
3
!3
!2

34. Counting Outcomes
0! is defined as being equal to 1. Let’s see why.
4
!4
!3
Writing this out using the definition of factorials.
4
1*2*3*4
4
1*2*3*4
!3
3
!3
!2
2
!2
!1

35. Counting Outcomes
0! is defined as being equal to 1. Let’s see why.
4
!4
!3
Writing this out using the definition of factorials.
4
1*2*3*4
4
1*2*3*4
!3
3
!3
!2
2
!2
!1
1
!1
!0so, the next logical conclusion is that

40. 9/11/2013 40
An unbiased die is tossed. Find the
probability of getting a
multiple of 3?
The possible options are : 1 to 6.
there are only 2 multiples of 3 : 3,6
so probability is (number of favourable
outcomes ) / (total number of possibilities)
= 2/6 = 1/3 answer

41. 9/11/2013 41
In a simultaneous throw of a pair of
dice,find the
probability of getting a total more
than 7?
We can have 36 possibilities (6*6) however, we need only
those cases where the total is 8 or more. These are :
(6,2),(6,3),(6,4),(6,5),(6,6),(5,3),(5,4),(5,5),(5,6),(4,4),(4,5),(4,6
),(3,5),(3,6),(2,6) =15
answer = 15/36 = 5/12 answer

42. 9/11/2013 42
A bag contains 6 white and 4 black
balls .Two balls are
drawn at random .Find the
probability that they are of the
same colour?
Both are white : 6/10*5/9
both are black = 4/10*3/9
add them : =42/90 or 7/15
or : 6c2/10C2*1/2 + 4c2/10c2
=21/45 = 7/15 answer

43. 9/11/2013 43
Two dice are thrown together.What
is the probability that the sum of the
number on
the two faces is divisible by 4 or 6?
The possibilities are : (1,3)(1,5) (2,2)
(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)
thus we are able to get 14 out of 36. so answer = 7/18
answer

44. 9/11/2013 44
Two cards are drawn at random
from a pack of 52 cards What
is the probability that either both
are black or both are
queens?
Both are black = 26/52 * 25/51=25/102
both are queens : 4/52 * 3/51=3/663
both are black queens : 2/52*1/51 = 1/1326
now add them : (25/102 + 3/663 – 1/1326)
=(325+6-1)/1326
=330/1326 or .25 answer

45. 9/11/2013 45
Two dices are tossed the probability
that the total score
is a prime number?
Prime numbers are : 1,2,3,5,7,11
totals are :
(1,2),(1,1),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4
),(4,1),(4,3),(5,2),(5,6),6,1),(6,5)
=15/36 answer

46. 9/11/2013 46
Two dice are thrown simultaneously .what is the
probability
of getting two numbers whose product is even?
If any one of the two numbers is an even number, the
product will be even number. Thus we should pick up all
those cases when both the numbers are odd numbers :
(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3) (5,5)
thus there are only 9 such cases. Remove them from 36,
we get : 27 cases
answer : 27/36 answer

47. 9/11/2013 47
In a lottery ,there are 10 prizes and 25 blanks.
A lottery is drawn at random.
what is the probability of getting a prize ?
10/(10+25)
=10/35 or 2/7 answer

48. 9/11/2013 48
In a class ,30 % of the students offered English,20 %
offered Hindi and 10 %offered Both.
If a student is offered at random, what is the probability
that he has offered English or Hindi?
30+20-10 = 40% or .4 answer

49. 9/11/2013 49
Two cards are drawn from a pack of 52 cards .What is the
probability that either both are Red or both are Kings?
Both are red ½ * 25/51
both are king = 4/52 + 3/51
now add both these answers =55/221

50. 9/11/2013 50
one card is drawn at random from a pack of 52 cards.What is
the probability that the card drawn is a face card?
Face cards are : Jack, queen, king
total = 12
12/52 answer

51. 9/11/2013 51
A man and his wife appear in an interview for two vacancies in
the same post.The probability of husband's selection is 1/7 and
the probabililty of wife's selection is 1/5.What is the probabililty
that only one of them is selected?
Husband + not wife
=1/7 * 4/5 = 4/35
wife + not husband
=1/5 * 6/7 = 6/35
add = 10/35 answer

52. 9/11/2013 52
From a pack of 52 cards,one card is drawn at
random.What is the probability that the card is a
10 or a spade?
4/52 + 13/52 – 1/52
=16/52 answer

53. 9/11/2013 53
A bag contains 4 white balls ,5 red and 6 blue balls .Three
balls are drawn at random from the bag.What is the probability
that all of them are red ?
5/15*4/14*3/13
or 5c2/15c2 =
=2/91

54. 9/11/2013 54
A box contains 10 block and 10 white balls.What is the
probability of drawing two balls of the same colour?
Both are black :
10/20 * 9/19 =9/38
+both are white :
10/20 * 9/19 =9/38
or
black : 10c2 / 20c2
+white : 10c2 / 20c 2
=90/190

55. 9/11/2013 55
A box contains 20 electricbulbs ,out of which 4 are
defective, two bulbs are chosen at random from this
box.What is the probability that at least one of these is
defective ?
In such questions (at least one type), it is better
to reverse the question, solve it and deduct the
answer from 1. So here we shall first calculate
the probability of getting no defective bulb.
Let us assume that no bulb is defective :
16/20 * 15/19 = 12/19
at least one is defective = 1 -12/19
= 7/19 answer

56. 9/11/2013 56
Two cards are drawn together from
apack of 52 cards.What is the
probability that
one is a spade and one is a heart ?
First is spade and 2nd heart :
13/52 * 13/51 = 13/204
First is heart and 2nd spade :
13/52 * 13/51 = 13/204
add them : 13/102 answer

57. 9/11/2013 57
The probability that a card drawn
from a pack of 52 cards will be a
diamond or a king?
13/52 + 4/52 – 1/52
=16/52