Probability Of An EventProbability Of An Event

Dhon G. Dungca, M.Eng’g.Dhon G. Dungca, M.Eng’g.

PROBABILITY PROBABILITY

provides a method by which we can evaluate the likelihood of the occurrence of events resulting from a statistical experiment. Connotes the “chance” or “likelihood” that something will happen or occur.

provides a method by which we can evaluate the likelihood of the occurrence of events resulting from a statistical experiment. Connotes the “chance” or “likelihood” that something will happen or occur.

PROBABILITY PROBABILITY

If an experiment can result in N equally likely outcomes of which nA of these outcomes correspond to event A, then the probability of event A is

P(A) = nA / N

If an experiment can result in N equally likely outcomes of which nA of these outcomes correspond to event A, then the probability of event A is

P(A) = nA / N

The simple axioms of probability The simple axioms of probability

Axiom I: 0 P(A) 1The probabilities of any event A are non-negative, real numbers lying between the values of zero and one, inclusive.

Axiom II: P(S) = 1The probability of the entire space is 1.

Axiom I: 0 P(A) 1The probabilities of any event A are non-negative, real numbers lying between the values of zero and one, inclusive.

Axiom II: P(S) = 1The probability of the entire space is 1.

The simple axioms of probability The simple axioms of probability

Axiom III: P(Ø) = 0The probability of the impossible event is 0.

Axiom IV: P(E1) = 1

The probability of the events in the sample space are added, the result should be unity.

Axiom III: P(Ø) = 0The probability of the impossible event is 0.

Axiom IV: P(E1) = 1

The probability of the events in the sample space are added, the result should be unity.

Example 1 Example 1

A coin is tossed twice. What is the probability that at least one head occurs?

S = {HH,HT,TH,TT}Let Event A = the event of at least 1

head

P(A) = nA = 3 N 4

A coin is tossed twice. What is the probability that at least one head occurs?

S = {HH,HT,TH,TT}Let Event A = the event of at least 1

head

P(A) = nA = 3 N 4

Example 2 Example 2

In rolling a balanced die, what are probabilities of getting a) 2 or 3?b) odd number

S = {1, 2, 3, 4, 5, 6}a) P(2U3) = 1/6 + 1/6 = 2/6 or 1/3b) P(odd) = 3/6 or 1/2

In rolling a balanced die, what are probabilities of getting a) 2 or 3?b) odd number

S = {1, 2, 3, 4, 5, 6}a) P(2U3) = 1/6 + 1/6 = 2/6 or 1/3b) P(odd) = 3/6 or 1/2

Example 3 Example 3

A box contains 50 marbles of different colors. 10 are red, 17 are blue, 20 are white & 3 are black. If one of these balls is drawn at random, what is the probability that it will bea) blackb) blue

c) red or white d) neither red nor blue

A box contains 50 marbles of different colors. 10 are red, 17 are blue, 20 are white & 3 are black. If one of these balls is drawn at random, what is the probability that it will bea) blackb) blue

c) red or white d) neither red nor blue

Example 4 Example 4

What is the probability of a) getting a queen?

b) not getting a queen?

a) P(Q) = 4/52 = 1/13b) P(Q’) = 52/52 – 4/52

= 48/52 or 12/13

What is the probability of a) getting a queen?

b) not getting a queen?

a) P(Q) = 4/52 = 1/13b) P(Q’) = 52/52 – 4/52

= 48/52 or 12/13

Example 5 Example 5

When a pair of balanced dice is thrown, what is the probability of getting a total sum of 9?

nsum9 = (6,3), (3,6), (4,5), (5,4)

N = (6)(6) = 36

P(sum9) = n9/N = 4/36 = 1/9

When a pair of balanced dice is thrown, what is the probability of getting a total sum of 9?

nsum9 = (6,3), (3,6), (4,5), (5,4)

N = (6)(6) = 36

P(sum9) = n9/N = 4/36 = 1/9

Example 6Example 6

In a poker hand consisting of 5 cards, find the probability of holding 2 aces and 3 jacks.P(2A3J) = n(2A3J) / N

n(2A3J) = (# of ways of being dealt 2 aces) (# of ways

of being dealt 3 jacks)N = total # of ways of a 5-card poker hands

In a poker hand consisting of 5 cards, find the probability of holding 2 aces and 3 jacks.P(2A3J) = n(2A3J) / N

n(2A3J) = (# of ways of being dealt 2 aces) (# of ways

of being dealt 3 jacks)N = total # of ways of a 5-card poker hands

Example 6Example 6

# of ways of being dealt 2 aces = 42

# of ways of being dealt 3 jacks = 43

total # of ways of a 5-card poker hands = 52 5

4 4

P(2A3J) = 2 3 52 5

P(2A3J) = 0.9 x 10-5

# of ways of being dealt 2 aces = 42

# of ways of being dealt 3 jacks = 43

total # of ways of a 5-card poker hands = 52 5

4 4

P(2A3J) = 2 3 52 5

P(2A3J) = 0.9 x 10-5

Example 7 Example 7

A pair of dice is tossed. What is the probability of gettinga) a 6?b) a total of 7?c) at most a total of 5?

a) P(6) = 11/36b) P(tot7) = 6/36 or 1/6c) P(tot5) = 10/36 or 5/18

A pair of dice is tossed. What is the probability of gettinga) a 6?b) a total of 7?c) at most a total of 5?

a) P(6) = 11/36b) P(tot7) = 6/36 or 1/6c) P(tot5) = 10/36 or 5/18

ADDITIVE RULES ADDITIVE RULES

Theorem 4.10If A & B are any two events, thenP(AB) = P(A) + P(B) – P(AB)

Theorem 4.10If A & B are any two events, thenP(AB) = P(A) + P(B) – P(AB)

ABA B

S

Example 8Example 8

What is the probability of getting a King or a Diamond or both in a single draw from a well-shuffled deck of playing cards?

P(KD) = P(K) + P(D) – P(KD) = 4/52 + 13/52 – 1/52 = 16/52 or 4/13

What is the probability of getting a King or a Diamond or both in a single draw from a well-shuffled deck of playing cards?

P(KD) = P(K) + P(D) – P(KD) = 4/52 + 13/52 – 1/52 = 16/52 or 4/13

Example 9Example 9

A candidate runs for offices A & B. He assesses the probability of winning A as 0.10 and winning B as 0.25. The probability of winning both is 0.05. What is the probability of winninga) at least 1 office?b) office b alone?

a) P(AB) = P(A) + P(B) – P(AB)b) P(B alone) = P(B) – P(AB)

A candidate runs for offices A & B. He assesses the probability of winning A as 0.10 and winning B as 0.25. The probability of winning both is 0.05. What is the probability of winninga) at least 1 office?b) office b alone?

a) P(AB) = P(A) + P(B) – P(AB)b) P(B alone) = P(B) – P(AB)

Example 10Example 10

The probability that a student passes mathematics is 2/3, and the probability that he passes English is 4/9. If the probability of passing at least one course is 4/5, what is the probability that he will pass both course?

The probability that a student passes mathematics is 2/3, and the probability that he passes English is 4/9. If the probability of passing at least one course is 4/5, what is the probability that he will pass both course?

Example 11Example 11

The members of a committee is composed of the following 5 persons:Person Sex Age

1 M 272 M 303 F 404 F 305 M 37

If the members of the committee decide to elect a chairperson by random draw, what is the probability that the chairperson is either male or over 35?

The members of a committee is composed of the following 5 persons:Person Sex Age

1 M 272 M 303 F 404 F 305 M 37

If the members of the committee decide to elect a chairperson by random draw, what is the probability that the chairperson is either male or over 35?

ADDITIVE RULES ADDITIVE RULES

Corollary 1If A & B are mutually exclusive, thenP(AB) = P(A) + P(B)

Corollary 1If A & B are mutually exclusive, thenP(AB) = P(A) + P(B)

A B

S

ADDITIVE RULES ADDITIVE RULES

Corollary 2If A1, A2, A3,…, An are mutually exclusive, thenP(A1A2A3…An) = P(A1) + P(A2) + P(A3) +…+ P(An)

Corollary 2If A1, A2, A3,…, An are mutually exclusive, thenP(A1A2A3…An) = P(A1) + P(A2) + P(A3) +…+ P(An)

Example 12Example 12

In drawing one card at random from a deck of 52 well-shuffled playing cards, what is the probability that you will get a King or a Queen?

P(KQ) = P(K) + P(Q) = 4/52 + 4/52 = 8/52 or 2/13

In drawing one card at random from a deck of 52 well-shuffled playing cards, what is the probability that you will get a King or a Queen?

P(KQ) = P(K) + P(Q) = 4/52 + 4/52 = 8/52 or 2/13

ADDITIVE RULES ADDITIVE RULES

Theorem 4.11If A & A’ complementary events, thenP(A) + P(A’) = 1

Theorem 4.11If A & A’ complementary events, thenP(A) + P(A’) = 1