Event – subset of the sample space of an experiment
Simple event – consists of exactly one outcome Compound event – two or more simple events Cardinality of a set A- number of elements in set A and is denoted by n(A)
Subset – Let E be an event of the sample space S. Since every outcome in E is an outcome in S, we say that event is a subset of S, denoted by E ⊂ S.
Union – denoted by A ∪ B, is the event
consisting of all sample points belonging to either A or B.
intersection - denoted by A ∩ B, is the event consisting of all sample points belonging to both A and B.
Complement - denoted by Ac, is the event consisting of all sample points which are not in A. P(A) + P(Ac ) = 1
Mutually exclusive events – cannot happen at the same time
The probability P of an event E is the ratio of favourable outcomes of an event E to the total number of possible outcomes.
COMPOUND EVENTS – combines at least two simple events, either the union of two simple events or the
intersection of two simple events Mutually Exclusive Events Compound Probability
P(A or B) = P(A) + P(B) or P(A ∪ B) = P(A) + P(B)
Not Mutually Exclusive Events Compound
Probability
P(A or B) = P(A) + P(B) - P(A and B)
P(A ∪ B) = P(A) + P(B) - P(A ⋂ B)
What is the probability of choosing a card that is a heart or an eight from a deck of cards?
P(A) = Probability of selecting a heart
¿ 13 52 = 1
4
P(B) = Probability of selecting an eight
¿ 4 52 = 1
13
P(A ∩ B) = Probability of selecting an eight and a heart
¿ 1
52
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
¿ 13 52 + 4
52 − 1 52 = 16
52 = 4 13
Independent events – does not affect the outcome of any other
events (P(A and B) = P(A) • P(B)
Examples: roll dice, coin flip, problems with replacement
Dependent event – does affect the
outcome of any other events (P(A and B) = P(A) • P(B following A)) Examples: deck of cards, selecting item from container, problems without replacement Example:
A coin and a six-sided die are tossed. Find the probability of
a. getting the tail When a coin is tossed, ther are only two possible outcomes, a head and a tail. Thus,
P (T )= 1 2
b. getting a even number
When a six-sided die is tossed, there are six possible outcomes (1, 2, 3,4 ,5, 6). There are 3 favorable outcomes (2, 4, 6,). Thus,
P (even)= 3 6 = 1
2
c. getting the tail and a even number
There are 12 possible outcomes. The list of the favorable outcomes is {T2, T4, T6}. Thus,
P (tail ∧an even number )= 3 12 = 1
4
If A and B are
independent events, find the indicated probability.
a.
P ( A )= 3 4 , P ( B)= 5
6 , P ( A ∩ B )=?
P ( A ∩ B )=P( A )⋅ P (B )
¿ 3 4 ⋅ 5
6
¿ 15 24 ∨ 5
8
b.P ( A )=0.5, P ( A ∩ B )=0.3, P ( B )=?
P ( A ∩ B )=P( A )⋅ P (B ) P ( B)= P ( A ∩ B )
P ( A )
¿ 0.3 0.5 ∨0.6
An urn contains 3 blue , 4 white, and 5 red marbles. Two marbles are drawn at random.
Find the probability of getting a red marble and a white mable in
each of the following conditions:
a. If 2 marbles are drawn with
replacement Drawing with
replacement means that the first marble is replaced before the second marble is drawn.If you are drawing with
replacement, then the two events (first draw and second draw) are independent.
P(red marble) = 5/12 P(white mable) = 4/12
= 1/3
P(one red and one white)
=
5 12 ∙ 1
3 + 1 3 ∙ 5
12 = 5 18
b. If 2 marbles aredrawn (one after the other) without replacement P(red then white marble) = P(red marble) • P(white marble on the second draw | red on first draw)
=
5 12 ∙ 4
11 = 20 132 = 5
33
P(white then red marble) = P(white marble) • P(red marbleon the second draw | white on first draw)
=
4 12 ∙ 5
11 = 20 132 = 5
33
Hence, P(one red and one white marble) = 5/33 + 5/33 = 10/33 A coin tossed and a die rolled. Find theprobability of landing on the head side of coin and rolling a 6 on the die.
P(head) = 1/2 P(6) 1/6
P(head and 6) =
1
2 ∙ 1 6 = 1
12
A die tossed twice. Find the probability of getting a 2 or 3 on the first toss and a 4, 5, or 6 on the second toss.
P(A) = P(2 or 3) = 1/6 + 1/6 = 2/6
P(B) = P(4, 5, or 6) = 1/6 + 1/6 + 1/6 = 3/6 P(A ∩ B) = P(A) • P(B)
¿ 2 6 ∙ 3
6 = 6 36 ∨ 1
6
A bag contains 6 red balls, 4 green balls, and 5 blue balls. One ball is taken from the bag, then replaced. Then
another ball is taken from the bag.
a. What is the probability of
getting a red ball on the first take and then a blue ball on the second?
Let :
R – event if getting a red ball
B – even of getting a blue
G – event of getting a green
P (R ∩B ) = P(R) • P(B)
P (R ∩B ) =
6
15 ∙ 5 15 = 2
15
Since, n(R) = 15, n(A)
= 6, n(G) = 5, n(C) = 4, therefore,
n(all balls) = n(R) + n(B) + n(G) = 6 + 5 + 4 = 15.
b. What is the
probability of getting a green on the first take and another green on the second?
P (G ∩ G ) = P(G) • P(G)
P (G ∩ G ) =
4
15 ∙ 4 15 = 16
225
A bag contains 6 red balls, 4 green balls, and 5 blue balls. One ball is taken from the bag, without replacement.
Then another ball is taken from the bag.
What is the probability of getting a green on the first take and another green on the second?
P (G ∩ G ) = P(G) • P(G)
P (G ∩ G ) =
4
15 ∙ 3 1 4 = 2
3 5
mutually exclusive:
P(A or B) - P(A)+P(B) not mutually exclusive:
P(A or B) = P(A)+P(B)- P(A and B)
Compound Probability Mutually Exclusive Events:
P(A ∪ B) = P(A) + P(B) Not Mutually Exclusive:
P(A ∪ B) = P(A) + P(B) - P(A ⋂ B)
Independent events:
P(A and B) = P(A) • P(B)
Dependent event:
P(A and B) = P(A) • P(B following A)
