Computing probability

1. COMPUTING
PROBABILITIES
Prepared by: Ms. Lynde Amor P. Casayas

2. Objectives:
a) identify the sample space and determine the
probability of an event
b) determine the probability of compound events
using the addition and multiplication rules
c) determine the conditional probability of an event
d) apply probability concepts in solving problems

3.  Probability – primarily concerned with predicting
chances, especially the occurrence of an
event
 Experiment – any probability activity that can be
infinitely repeated and yield results
 Sample space – the set of all the possible outcomes
 Event – a subset of the sample space
Definition of Terms:

4.  is the numerical measure of the likelihood that an event will occur
 calculated as
where nis the number of the elements in the event and
nis the number of the elements in the sample space
 n(E) ≤ n(S)
 therefore,
Probability of an Event, denoted as

5. Lesson 2.1

6. Two coins are tossed
simultaneously. find the
probability of getting both
heads.
Sample space:
S = { HH, HT, TH, TT }
Example 1:
𝑃 ( 𝐸)=
𝑛 ( 𝐸)
𝑛( 𝑆)

7. Two coins are tossed
simultaneously. find the
probability of getting both
heads.
Sample space:
S = { HH, HT, TH, TT }
Example 1:
¿
¿¿
𝟒
¿
𝑃 ( 𝐸)=
𝑛 ( 𝐸)
𝑛( 𝑆)

8. Two coins are tossed
simultaneously. find the
probability of getting both
heads.
Sample space:
S = { HH, HT, TH, TT }
Example 1:
¿
𝟏
𝟒
¿ 0.25
¿25 %
𝑃 ( 𝐸)=
𝑛 ( 𝐸)
𝑛( 𝑆)

9. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 2:
A pair of dice is thrown. What is the probability of
getting doubles?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

10. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 2:
A pair of dice is thrown. What is the probability of
getting doubles?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

11. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 2:
A pair of dice is thrown. What is the probability of
getting doubles?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

12. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 3:
A pair of dice is thrown. What is the probability of
getting a sum of 5?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

13. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 3:
A pair of dice is thrown. What is the probability of
getting a sum of 5?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

14. 𝑺=¿
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Example 3:
A pair of dice is thrown. What is the probability of
getting a sum of 5?
Sample space:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

15. 𝑺=¿
Example 4:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking a
black jack?
Sample space:

16. 𝑺=¿
Example 4:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking a
black jack?
Sample space:

17. 𝑺=¿
Example 4:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking a
black jack?
Sample space:

18. 𝑺=¿
Example 5:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking an
ace?
Sample space:

19. 𝑺=¿
Example 5:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking an
ace?
Sample space:

20. 𝑺=¿
Example 5:
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
If a card is selected at random from a standard
deck of 52 cards. Find the probability of picking an
ace?
Sample space:

21. Example 6:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

22. Example 6:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

23. Example 6:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

24. Example 7:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
not getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

25. Example 7:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
not getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

26. Example 7:
Oliver has a bag with 6 red, 4 blue, and 8 green
marbles.
a. What is the probability of
not getting a red marble?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

27. Example 8:
A four-digit number is
formed using the digits
1, 2, 3, 5, 8, and 9
What is the probability
that the number formed is
even?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

28. Example 8:
A four-digit number is
formed using the digits
1, 2, 3, 5, 8, and 9
What is the probability
that the number formed is
even?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
¿
¿¿ ∗¿ ∗¿
𝟔∗𝟔∗𝟔∗𝟔

29. Example 8:
A four-digit number is
formed using the digits
1, 2, 3, 5, 8, and 9
What is the probability
that the number formed is
even?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
¿
𝟔∗𝟔∗𝟔∗𝟐
𝟔∗𝟔∗𝟔∗𝟔

30. Example 8:
A four-digit number is
formed using the digits
1, 2, 3, 5, 8, and 9
What is the probability
that the number formed is
even?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
¿
𝟔∗𝟔∗𝟔∗𝟐
𝟔∗𝟔∗𝟔∗𝟔

31. Example 8:
A four-digit number is
formed using the digits
1, 2, 3, 5, 8, and 9
What is the probability
that the number formed is
even?
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
¿
𝟔∗𝟔∗𝟔∗𝟐
𝟔∗𝟔∗𝟔∗𝟔

32. Twenty people with two
sisters among them want to
sit around a round table
such that the two sisters sit
together. What is the
probability that the two
sisters would sit together?
Example 9:
¿𝟐!∗¿¿
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

33. Twenty people with two
sisters among them want to
sit around a round table
such that the two sisters sit
together. What is the
probability that the two
sisters would sit together?
Example 9:
¿
𝟐!∗(𝟏𝟗−𝟏)!
(𝟐𝟎−𝟏)!
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

34. Twenty people with two
sisters among them want to
sit around a round table
such that the two sisters sit
together. What is the
probability that the two
sisters would sit together?
Example 9:
¿
𝟐!∗(𝟏𝟗−𝟏)!
(𝟐𝟎−𝟏)!
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

35. Twenty books consisting
of 7 algebra, 4 geometry, and
9 trigonometry books are to
be randomly arranged on a
shelf. What is the probability
that the geometry books are
arranged next to one
another?
Example 10:
¿
𝟒 ! ∗¿
𝟐𝟎 !
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

36. Twenty books consisting
of 7 algebra, 4 geometry, and
9 trigonometry books are to
be randomly arranged on a
shelf. What is the probability
that the geometry books are
arranged next to one
another?
Example 10:
¿
𝟒!∗𝟏𝟕!
𝟐𝟎!
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)

37. Twenty books consisting
of 7 algebra, 4 geometry, and
9 trigonometry books are to
be randomly arranged on a
shelf. What is the probability
that the geometry books are
arranged next to one
another?
Example 10:
¿
𝟒!∗𝟏𝟕!
𝟐𝟎!
𝑃 ( 𝐸)=
𝑛(𝐸)
𝑛(𝑆)
¿
𝟏
𝟐𝟖𝟓

38.  finding the probability of an event followed by
another event, whether from the same experiment
or from another
 individual probabilities are calculated and then
multiplied
 this rule especially applies when the events are
independent – the outcome of one does not affect
the other
Probability Involving a Sequence of Events

39.  two events A and B are independent when the outcome
of the first does not affect the outcome of the other
 the probability of event A followed by another
independent event B occurring is given by
Multiplication Rule of Probability: Independent Events
Lesson 2.2

40. Example 11:
A coin is tossed and then a die is
rolled. What is the probability of getting
a head followed by a 4?
𝑃 ( 𝐴∩ 𝐵)=
¿¿
2
∗
¿¿
6
¿ ¿
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)

41. Example 11:
A coin is tossed and then a die is
rolled. What is the probability of getting
a head followed by a 4?
𝑃 ( 𝐴∩ 𝐵)=
1
2
∗
1
6
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)

42. Example 11:
A coin is tossed and then a die is
rolled. What is the probability of getting
a head followed by a 4?
𝑃 ( 𝐴∩ 𝐵)=
1
2
∗
1
6
¿
𝟏
𝟏𝟐
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)

43. Example 12:
A pair of dice is rolled thrice. What is
the probability of getting doubles in the
three rolls?
𝑃 ( 𝐴∩ 𝐵 ∩𝐶)=
¿¿
36
∗
¿¿
36
∗
¿¿
36
¿¿¿

44. Example 12:
A pair of dice is rolled thrice. What is
the probability of getting doubles in the
three rolls?
𝑃 ( 𝐴∩ 𝐵∩𝐶)=
6
36
∗
6
36
∗
6
36

45. Example 12:
A pair of dice is rolled thrice. What is
the probability of getting doubles in the
three rolls?
𝑃 ( 𝐴∩ 𝐵∩𝐶)=
6
36
∗
6
36
∗
6
36
¿
𝟏
𝟐𝟏𝟔

46. Example 13:
There are 7 green, 4 blue, and 9 red
marbles in a box. A marble is drawn at
random and, after recording its color, the
marble is returned in the box. Another
marble is then drawn. What is the
probability that both marbles are red?
𝑃 ( 𝐴∩ 𝐵)=
¿¿
20
∗
¿¿
20
¿ ¿
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)

47. Example 13:
There are 7 green, 4 blue, and 9 red
marbles in a box. A marble is drawn at
random and, after recording its color, the
marble is returned in the box. Another
marble is then drawn. What is the
probability that both marbles are red?
𝑃 ( 𝐴∩ 𝐵)=
9
20
∗
9
20
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)

48. Example 13:
There are 7 green, 4 blue, and 9 red
marbles in a box. A marble is drawn at
random and, after recording its color, the
marble is returned in the box. Another
marble is then drawn. What is the
probability that both marbles are red?
𝑃 ( 𝐴∩ 𝐵)=
9
20
∗
9
20
𝑃 ( 𝐴∩ 𝐵)=𝑃( 𝐴)∗𝑃( 𝐵)
¿
𝟖𝟏
𝟒𝟎𝟎

49.  if events A and B are dependent events such that the
outcome of A affects the outcome of B, then the
probability that “event A followed by event B” happens is
 where P(A) is the probability that event A happens and
P(B|A) is the probability that event B happens after A
happened
Multiplication Rule of Probability: Dependent Events
Lesson 2.3

50. Example 14:
Jacob draws 2 cards at random
from a standard deck of 52 cards
without replacement. What is the
probability of getting 2 ace cards?
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

51. Example 14:
Jacob draws 2 cards at random
from a standard deck of 52 cards
without replacement. What is the
probability of getting 2 ace cards?
𝑃 ( 𝐴∩ 𝐵)=
¿¿
52
∗
¿¿
51
¿¿
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

52. Example 14:
Jacob draws 2 cards at random
from a standard deck of 52 cards
without replacement. What is the
probability of getting 2 ace cards?
𝑃 ( 𝐴∩ 𝐵)=
4
52
∗
3
51
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

53. Example 14:
Jacob draws 2 cards at random
from a standard deck of 52 cards
without replacement. What is the
probability of getting 2 ace cards?
𝑃 ( 𝐴∩ 𝐵)=
4
52
∗
3
51
¿
12
2652
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

54. Example 14:
Jacob draws 2 cards at random
from a standard deck of 52 cards
without replacement. What is the
probability of getting 2 ace cards?
𝑃 ( 𝐴∩ 𝐵)=
4
52
∗
3
51
¿
𝟏
𝟐𝟐𝟏
¿
12
2652
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

55. Example 15:
Gabriel has a candy box containing 9
chocolate candies and 12 lollipops. If he
chooses 2 of them at random, what is the
probability that he gets 2 lollipops?
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

56. Example 15:
Gabriel has a candy box containing 9
chocolate candies and 12 lollipops. If he
chooses 2 of them at random, what is the
probability that he gets 2 lollipops?
𝑃 ( 𝐴∩ 𝐵)=
¿¿
21
∗
¿¿
20
¿¿
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

57. Example 15:
Gabriel has a candy box containing 9
chocolate candies and 12 lollipops. If he
chooses 2 of them at random, what is the
probability that he gets 2 lollipops?
𝑃 ( 𝐴∩ 𝐵)=
12
21
∗
11
20
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

58. Example 15:
Gabriel has a candy box containing 9
chocolate candies and 12 lollipops. If he
chooses 2 of them at random, what is the
probability that he gets 2 lollipops?
𝑃 ( 𝐴∩ 𝐵)=
12
21
∗
11
20
¿
𝟏𝟑𝟐
𝟒𝟐𝟎
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

59. Example 15:
Gabriel has a candy box containing 9
chocolate candies and 12 lollipops. If he
chooses 2 of them at random, what is the
probability that he gets 2 lollipops?
𝑃 ( 𝐴∩ 𝐵)=
12
21
∗
11
20
¿
𝟏𝟏
𝟑𝟓
¿
𝟏𝟑𝟐
𝟒𝟐𝟎
𝑃 ( 𝐴∩𝐵)=𝑃( 𝐴)∗𝑃(𝐵∨𝐴)

60.  the probability of a single event that is made up
of two different mutually exclusive events is
given by
Addition Rule of Probability: Mutually Exclusive Events
Lesson 2.4

61. Example 16:
A card is drawn at random from a
standard deck of cards. What is the
probability that the card drawn is an
ace or a king?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟓𝟐
+
¿¿
𝟓𝟐

62. Example 16:
A card is drawn at random from a
standard deck of cards. What is the
probability that the card drawn is an
ace or a king?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟒
𝟓𝟐

63. Example 16:
A card is drawn at random from a
standard deck of cards. What is the
probability that the card drawn is an
ace or a king?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟒
𝟓𝟐
¿
𝟖
𝟓𝟐

64. Example 16:
A card is drawn at random from a
standard deck of cards. What is the
probability that the card drawn is an
ace or a king?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟒
𝟓𝟐
¿
𝟖
𝟓𝟐
¿
𝟐
𝟏𝟑

65. Example 17:
A pair of dice is rolled. What is the
probability that the sum is 5 or 6?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟑𝟔
+
¿¿
𝟑𝟔

66. Example 17:
A pair of dice is rolled. What is the
probability that the sum is 5 or 6?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟑𝟔
+
𝟓
𝟑𝟔

67. Example 17:
A pair of dice is rolled. What is the
probability that the sum is 5 or 6?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟑𝟔
+
𝟓
𝟑𝟔
¿
𝟗
𝟑𝟔

68. Example 17:
A pair of dice is rolled. What is the
probability that the sum is 5 or 6?
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟑𝟔
+
𝟓
𝟑𝟔
¿
𝟗
𝟑𝟔
¿
𝟏
𝟒

69. A card is selected at random from
a standard deck of 52 cards. What is
the probability that it is a red king or a
black queen?
Example 18:
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟓𝟐
+
¿¿
𝟓𝟐

70. A card is selected at random from
a standard deck of 52 cards. What is
the probability that it is a red king or a
black queen?
Example 18:
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟐
𝟓𝟐
+
𝟐
𝟓𝟐

71. A card is selected at random from
a standard deck of 52 cards. What is
the probability that it is a red king or a
black queen?
Example 18:
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟐
𝟓𝟐
+
𝟐
𝟓𝟐
¿
𝟒
𝟓𝟐

72. A card is selected at random from
a standard deck of 52 cards. What is
the probability that it is a red king or a
black queen?
Example 18:
𝑷 ( 𝑨⋃ 𝑩)=𝑷 ( 𝑨)+ 𝑷( 𝑩)
𝑷 ( 𝑨⋃ 𝑩)=
𝟐
𝟓𝟐
+
𝟐
𝟓𝟐
¿
𝟒
𝟓𝟐
¿
𝟏
𝟏𝟑

73.  if events A and B are given such that A and B
have common outcomes, then the probability
of the event “A” or “B” is determined by
-
Addition Rule of Probability: Inclusive Events
Lesson 2.5

74. If a card is selected at random from
a standard deck of cards, what is the
probability that the card drawn is a red
card or a face card?
Example 19:
-
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟓𝟐
+
¿¿
𝟓𝟐
−
¿¿
𝟓𝟐

75. If a card is selected at random from
a standard deck of cards, what is the
probability that the card drawn is a red
card or a face card?
Example 19:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟏𝟐
𝟓𝟐
−
𝟔
𝟓𝟐

76. If a card is selected at random from
a standard deck of cards, what is the
probability that the card drawn is a red
card or a face card?
Example 19:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟏𝟐
𝟓𝟐
−
𝟔
𝟓𝟐
¿
𝟑𝟐
𝟓𝟐

77. If a card is selected at random from
a standard deck of cards, what is the
probability that the card drawn is a red
card or a face card?
Example 19:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟏𝟐
𝟓𝟐
−
𝟔
𝟓𝟐
¿
𝟑𝟐
𝟓𝟐
¿
𝟖
𝟏𝟑

78. A single card is selected from a
deck of cards. Find the probability
that it is a king or a red card?
Example 20:
-
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟓𝟐
+
¿¿
𝟓𝟐
−
¿¿
𝟓𝟐

79. A single card is selected from a
deck of cards. Find the probability
that it is a king or a red card?
Example 20:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟐𝟔
𝟓𝟐
−
𝟐
𝟓𝟐

80. A single card is selected from a
deck of cards. Find the probability
that it is a king or a red card?
Example 20:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟐𝟔
𝟓𝟐
−
𝟐
𝟓𝟐
¿
𝟐𝟖
𝟓𝟐

81. A single card is selected from a
deck of cards. Find the probability
that it is a king or a red card?
Example 20:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟒
𝟓𝟐
+
𝟐𝟔
𝟓𝟐
−
𝟐
𝟓𝟐
¿
𝟐𝟖
𝟓𝟐
¿
𝟕
𝟏𝟑

82. If a card is selected at random
from a standard deck of cards, what
is the probability that the card drawn
is a black card or an ace card?
Example 21:
-
𝑷 ( 𝑨⋃ 𝑩)=
¿¿
𝟓𝟐
+
¿¿
𝟓𝟐
−
¿¿
𝟓𝟐

83. If a card is selected at random
from a standard deck of cards, what
is the probability that the card drawn
is a black card or an ace card?
Example 21:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟒
𝟓𝟐
−
𝟐
𝟓𝟐

84. If a card is selected at random
from a standard deck of cards, what
is the probability that the card drawn
is a black card or an ace card?
Example 21:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟒
𝟓𝟐
−
𝟐
𝟓𝟐
¿
𝟐𝟖
𝟓𝟐

85. If a card is selected at random
from a standard deck of cards, what
is the probability that the card drawn
is a black card or an ace card?
Example 21:
-
𝑷 ( 𝑨⋃ 𝑩)=
𝟐𝟔
𝟓𝟐
+
𝟒
𝟓𝟐
−
𝟐
𝟓𝟐
¿
𝟐𝟖
𝟓𝟐
¿
𝟕
𝟏𝟑

86.  the conditional probability of an event is the
probability of the event given the condition that
another event has previously occurred. It is
computed using the formula
Conditional Probability of an Event
Lesson 2.6

87. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

88. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝟐𝟓 %
𝟓𝟎 %
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

89. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝟐𝟓 %
𝟓𝟎 %
¿
𝟎.𝟐𝟓
.𝟓𝟎
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

90. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝟐𝟓 %
𝟓𝟎 %
¿
𝟏
𝟐
¿
𝟎.𝟐𝟓
.𝟓𝟎
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

91. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝟐𝟓 %
𝟓𝟎 %
¿
𝟏
𝟐
¿
𝟎.𝟐𝟓
.𝟓𝟎
¿𝟎.𝟓
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

92. A math teacher gave her class two test,
25% of the class passed both tests and 50%
of the class passed the first test. What
percent of those who passed the first test
also passed the second test?
Example 22:
𝑷 (𝑩∨𝑨)=
𝟐𝟓 %
𝟓𝟎 %
¿
𝟏
𝟐
¿𝟓𝟎%
¿
𝟎.𝟐𝟓
.𝟓𝟎
¿𝟎.𝟓
𝑷 (𝑩∨𝑨)=
𝑷 ( 𝑨∩𝑩)
𝑷 ( 𝑨)
𝒘𝒉𝒆𝒓𝒆 𝑷 ( 𝑨)≠𝟎

93. The probability that it is Friday and that a
student is absent is 0.03. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that a
student is absent given that today is Friday?
Example 23:

94. The probability that it is Friday and that a
student is absent is 0.03. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that a
student is absent given that today is Friday?
Example 23:
𝑷 (𝑩∨𝑨)=
𝟎.𝟎𝟑
𝟎.𝟐

95. The probability that it is Friday and that a
student is absent is 0.03. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that a
student is absent given that today is Friday?
Example 23:
𝑷 (𝑩∨𝑨)=
𝟎.𝟎𝟑
𝟎.𝟐
¿
𝟑
𝟐𝟎

96. The probability that it is Friday and that a
student is absent is 0.03. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that a
student is absent given that today is Friday?
Example 23:
𝑷 (𝑩∨𝑨)=
𝟎.𝟎𝟑
𝟎.𝟐
¿𝟎.𝟏𝟓
¿
𝟑
𝟐𝟎

97. THANK YOU!