CH 2
Probability
Sample Space and Events
Sample Space (S) = the set of all possible outcomes
Event (E) = a set consisting of possible outcomes of the experiment
Ex 1 If the outcome of an experiment consists of the determination of the sex of a newborn child, then
S = {m, f}
Event E1 = {f} denotes an event that the newborn child is female.
Event E2 = {m} denotes an event that the newborn child is male.
Ex 2 If an experiment involves horse racing (7 horses in total), and the outcome is any number from 1 to 7.
S = {any ordered arrangement from 1 to 7}
Event F = (2, 3, 1, 6, 5, 4, 7} denotes an event that the second horse wins
Ex 3 If we want to find the dosage of medicine that can successively cure patients, then
S = {0, }
Union
E F = all outcomes that are either in E or in F or both in E and F
E F can be written as E or F
Intersection
E F = all outcomes that are in both E and F
E F can be written as E and F
Ex 4 In Example 1, if E1 = {f} , E2 = {m}, then
E1 E2 = {f,m}
Ex 5 In Example 2, if A = {all outcomes that begin with 6}, and
B = {the 6th horse is at the second place}
A B = Event that the 6th horse win the 1st or 2nd place
Ex 6 In Example 3, if E1 = (0, 5) denotes an event that the amount of drug is less than 5 units, and E2 denotes an event that the amount of drug is between 2 and 10 units
E1 E2 = (2, 5) denotes an event that the amount of drug is between 2 and 5
Mutually Exclusive Events (Disjoint events)
Two events (A and B) are mutually exclusive if they cannot occur at the same time.
If two events are disjoint, then the probability of them both occurring at the same time is 0.
Mutually exclusive events: P(A B) = 0Ex 7 In Example 2, if A = {all outcomes that end with 5} = the 5th horse loses
B = {all outcomes that begin with 5} = the 5th horse wins
then A and B are mutually exclusive events?
Complement
consists of all outcomes in the sample space S that are not in E
Ex 8 In Example 1, if E1 = {f} , then = {m}
Ex 9 =
= A
Subset E1 E2 If all of the outcomes in E1 are also in E2
= E2 E1
If E1 E2 and E2 E1 then E1 = E2
Note We can also define ‘union’ and ‘intersection’ for more than 2 events.
E1 E2 . . . En
Or written as
Venn’s Diagram
E1 E2 E1 E2
Ex 10 A factory produces 3 types of switch gear which can be used with manual(M), semiautomatic, (S), andautomatic (A) machines. In testing 100 switches, the following results are found.
Switch / Number of switchesM, S, A
M, S
A
S,A
M / 10
25
5
15
the others
(a)Let M, S, A be the set of switch gears that can be used with different types of machines. Draw a Venn’s diagram and identify the number of switches that can be used with each machine.
(b)Find the number of switch gear in, S และ
Solution
Operation of Sets
Commutative law E1 E2 = E2 E1
E1 E2 = E2 E1
Associative law(E1 E2 ) E3 = E1 (E2 E3 )
(E1 E2 ) E3 = E1 (E2 E3 )
Distributive Law(E1 E2 ) E3 = (E1 E3 ) (E2 E3 )
(E1 E2 ) E3 = (E1 E3 ) (E2 E3 )
De Morgan’s Law(E1 E2 )’ =
(E1 E2 )’ =
Venn’s diagram can be used to prove the above laws, for example
(E1 E2 ) E3 = (E1 E3 ) (E2 E3 )
(a) E1 E3(b) E2 E3(c ) (E1 E2 ) E3
พิสูจน์ได้ว่า(E1 E2 ) E3
= (E1 E3 ) (E2 E3 )
Definition of Probability
If an experiment can result in any one of N outcomes that are equally likely to occur (sample space S = {1, 2,…, N}), and if event E consists of n outcomes,
then the probability of event E is:
P (E) = n / Nn = number of outcomes in event E, N = number of outcomes in the sample space
Probability Rules
1.Probability of an event E can be written as P (E)
0 P (E) 12.
P (S) = N / N = 13.
If E1 , E2 , ... , En are mutually exclusive (that is, events for which when I≠ j )
= where n = 1, 2, …,
Or P (E1 E2 . . . En ) = P (E1) + P (E2) + . . . + P (En)
Meaning : For any set of mutually exclusive events, the probability that at least one of these events occurs is equal to the sum of their respective probabilities.
4.
P (A’) = 1 – P (A)1 = P (S) fromAxiom 2
= P (A A’)
= P (A) + P (A’) fromAxiom 3
5.
P (A B) = P (A) + P (B) - P (A B)Proof The key idea is that, in adding P (A) and P (B), the probability of the intersection A B is actually counted twice, so P (A B) must be subtracted out..
It can also be proved by using Venn’s Diagram.
I, II, III are mutually exclusive events, thus
P (E1 E2 ) = P (I) + P (II) + P (III) …………………… (1)
P (E1 )= P (I) + P (II)
P (I )= P (E1 ) - P (II) …………………………………..(2)
P (E2 )= P (II) + P (III)
P (III )= P (E2 ) - P (II) …………………………………..(3)
Replace (2), (3) in(1) P (E1 E2 ) = P (E1 ) + P (E2 ) - P (II)
= P (E1 ) + P (E2 ) - P (E1 E2 )
The probability of a union of more than 2 events can be computed analogously. For 3 events:
P (A B C) = P (A) + P (B) + P (C) - P (A B) - P (A C) – P (B C) + P (A B C)A B C
The above law can be proved byVenn’s Diagram. WhenP(A), P(B), P(C) are added together, we have includedintersectionsof each event twice so we need to subtract them out. However, this has resulted in P (A B C)being subtracted more than once.
If A and B are mutually exclusive, then P (A B) = 0Ex 11 (From Devore, 2000)
In a certain residential suburb, 60% of all households subscribe to the metropolitan newspaper published in a nearby city, 80% subscribe to the local afternoon paper, and 50% of all households subscribe to both papers. If a household is selected at random, what is the probability that it subscribes to (1) at least one of the two newspapers and (2) exactly one of the two newspapers?
Sol. Let A = {subscribes to the metropolitan paper}
B = {subscribes to the local paper}
Given P (A) = 0.6, P (B) = 0.8, P (A B) = 0.5
(1). P (subscribes to at least one of the two newspapers) = P (A B)
= P (A) + P (B) - P (A B)
= 0.6 + 0.8 - 0.5 = 0.9
P (A B) = P (A) + P (A’B)
(2) From above0.9 = P (A B)
= P (A) + P (A’B)
= 0.6 + P (A’B)
P (A’B)= 0.3
Similarly,0.9 = P (A B)
= P (B) + P (A B’)
= 0.8 + P (A B’)
P (A B’)= 0.1
P (exactly one) = P (A B’) + P (A’B)
= 0.1 + 0.3 = 0.4
Ex 12 (From Devore, 2000) When 2 dices are rolled separately, there are N = 36 outcomes. If both the dices are fair, find the probability of getting the sum of both dices equal to 7.
Sol. S = (1,1), (1,2),…, (1,6), (2,1),..,(2,6),(3,1)…….(6,6) = 36 outcomes
A = {the sum of two numbers = 7}
= (1, 6), (2, 5), (3,4), (4,3), (5,2), (6,1)
P (A) = 6 / 36
Ex 13 (From Devore, 2000) In rolling a die that is made imbalanced so that the even sides are twice more likely to occur than odd sides.
(a) Find the probability that the die gives an even number
(b) Find the probability that the die gives a number that is not greater than 3
Sol. LetEi = the die gives a number i P (E1) = P (E3) = P (E5) = 1/9
P (E2) = P (E4) = P (E6) = 2/9
P (the die gives an even number) = P (E2) + P (E4) + P (E6) = 2/9 + 2/9 + 2/9 = 6/9
P (the die gives a number 3) = P (E1) + P (E2) + P (E3)= 1/9 + 2/9 + 1/9 = 4/9
Specific Addition Rule
Only valid when the events are mutually exclusive.
P(A B) = P(A) + P(B)2-1