Basic Principle of Counting

An ordered r-tuple of elements of a nonempty set Z, say (z₁, z₂,...,z_k), with distinct components (that is, z_i  zj) is called a permutation of r elements of Z. If set Z contains n

distinct elements, then the number of r-permutations of set Z is denoted by P(n,r) or nPr

A subset {z₁, z2,...,zk}} with k distinct elements of a nonempty set Z, is called a combination

of k elements of Z. If set Z contains n distinct elements, then the number of r-combinations of set Z is denoted by C(n,r) or ⬚n

r

4. If a multiple-choice test consists of 5 questions each with 4 possible answers of which only 1 is correct,

a. How many different ways can a student answer the questions?

A = {( x1, x2, x3, x4, x5): xi {a, b, c, d }, i= 1, 2, 3, 4, 5}. n(A) = (4)(4)(4)(4)(4) = 45 = 1024

1. How many different ways can a student answer all the 5 questions incorrectly?

B = {( x1, x2, x3, x4, x5): xi {a, b, c, d } – {y}, where y is the correct answer, i= 1, 2, 3, 4, 5}.

n(B) = (3)(3)(3)(3)(3) = 35 = 243

Exercise:

1. How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 are for numbers? What about if no letter or no number can be repeated in a single license plate?

2. Paul, John, Ringo and George have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if Paul and John can play all 4 instruments but George and Ringo can play only the piano and drums?

3. How many different ways can a true-false test consisting of 10 questions be answered?

Example: Suppose Z = {1, 2, 3, 4, 5}. List down all the possible permutations of 3 elements of Z. List down all possible combination of 3 elements of Z.

There are 60 possible permutations for this scenario.

Whereas, there are only 10 possible combinations as follows:

{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}.

6.4.2 Permutations and Combinations

If we are counting the number of ways k objects could be chosen from n objects without regards on being distinct (getting th e same object more than once). We could get it by the formula nk_.

While if we need to take into account that the selected objects needs to be distinct. We could get it by the formula n! = n x (n-1) x (n-2) x ... x 1. We also define 0! = 1.

Remarks:

There are k! distinct permutations associated with each combination. k! represents the number of ways in which you can arrange the elements that were included in the combination.

When we are counting the number of different groups of k objects that can be formed from n distinct objects, we would count the combinations while if we are interested in determining the number of different ordered arrangements of k objects selected from n distinct objects, we count the permutations.

Example:

1. Consider the experiment of tossing 4 distinguishable dice. a. How many possible outcomes are there?

 =

{

(x1, x2, x3, x4): xi Z = {1, 2, 3, 4, 5, 6}

}

n() = nk = 64 = 1296

b. How many possible outcomes are there for which no two dice show the same number of spots?

A = {(x1, x2, x3, x4): xi Z = {1, 2, 3, 4, 5, 6}, xi  xj for i j}

n(A)= (n)k = (6)4= (6)(5)(4)(3) =360

c. How many possible outcomes are there for which all the spots are even? B =

{

(x1, x2, x3, x4): xi Z = {2, 4, 6}

}

n(B) = nk = (3)4 = 81

2. From a group of 5 men and 7 women.

1. How many committees of 5 persons can be formed?

Since we are counting the number of different groups of k=5 objects that could be formed from a total of n=12 objects then we are counting the number of

2. How many committees of 2 men and 3 women can be formed?

The experiment can be divided into 2 stages: (a) the selection of the men and (b) the selection of women. Then by the basic principle of counting, there are n1n2

possible committees where n1= number of ways that the men can be selected

and n2= number of ways that the women can be selected.

( ) ( ). Thus, the number of committees consisting of 2 men and 3 women that can be formed is n1n2= ( ) ( )=(10)(35) = 350.

Exercise:

1. Consider the game of poker where a player is given 5 cards. a. How many 5-card poker hands are there?

b. How many of these 5-card poker hands contain exactly 3 hearts?

2. A class consists of 10 men and 20 women. An examination is given, and the students are ranked according to their performance. Assume that no two students obtain the same score.

a. How many different ranking are possible?

b. If the men were ranked just among themselves and the women among themselves, how many different ranking are possible?

3. Five separate awards are to be presented to selected students from a class of 30. How many different outcomes are possible if

a. A student can receive any number of awards; b. Each student can receive at most 1 award

Example: How many different letter arrangements can be formed using the letters P E P P E R?

There are

=

possible letter arrangements where n = number of letters r1 = number of P’s

r2 = number of E’s r3 = number of R’s

Additional Counting Theorems:

The number of permutations of n distinct objects in a circle is (n-1)!

The number of distinct permutations of n things of which r₁ are of the 1st _{kind, r}

2 are of

the 2nd _{kind, ..., r}

k are of the kth kind is

Take note that there are actually 6! permutations of the letters P1, E1, P2, P3, E3, R when

the 3 P’s and 2 E’s can be distinguished from each other. But in this case, we know that the letters cannot be distinguished from each other.

Exercise:

1. How many different signals, each consisting of 9 flags hung in a line, can be made from a set of 4 white flags, 3 red flags, and 2 blue flags, if all flags of the same color are identical?

2. How many ways can 20 new applicants be assigned to the 5 committees of an organization so that each committee will get 4 new applicants each?

Suppose an urn contains M balls, labeled 1 to M, and a sample of size n is drawn, then there are:

i. Mn ordered samples with replacement ii. (M)n ordered samples without replacement

iii. ( ) unordered samples without replacement

Example: Suppose an urn contains 10 balls, labeled 1 to 10, and a sample of size 5 is drawn.

1. How many ordered samples with replacement can be drawn? (10)5 2. How many ordered samples without replacement can be drawn? (10)5

3. How many unordered samples with replacement can be drawn? ( )

6.4.3 Special Results in Counting

Theorem:

Suppose an urn contains M balls, labeled 1 to M, and those labeled from 1 to K (K<M) are defective. Define A_k = the set containing all possible ordered samples of size n which contains k defectives.

(i) under sampling with replacement, n(A_k) =

Example:

1. Suppose an urn contains 10 balls, labeled 1 to 10, and those labeled from 1 to 4 are defective. Define A2 = the set containing all possible ordered samples of size 5 which

contains 2 defectives.

a. How many elements belong to A2 if sampling is done with replacement?

( ) . / ( )n _{( * ( )}

b. How many elements belong to A2 if sampling is done without replacement?

( ) . / ( ) ( *( ) ( )

2. If a multiple-choice test consists of 10 questions each with 4 possible answers such that a. there are exactly 7 correct answers

( ) . / ( )n _{( * ( )}

b. there are at least 7 correct answers

( ) ∑ 0. / ( )n ₁ ∑ [( * ( ) _]

c. there are at most 7 correct answers

( ) ∑ 0. / ( )n ₁

∑ [( * ( ) _]

Exercise: Suppose a group of 20 undergraduate students and 10 post graduate students are

available to fill certain student government posts. If 6 students are to be randomly selected from this group,

1. How many possible ordered samples with replacement are there; 2. How many possible ordered samples without replacement are there;

3. How many possible ordered samples with replacement will contain exactly 3 undergraduate students;

4. How many possible ordered samples without replacement will contain exactly 3 undergraduate students?

Bonus Exercise:

1. How many different letter arrangements can be made from the letters of the word M I S S I S S I P P I?

2. How many ways can 8 people be seated in a row if

a. there are no restrictions on the seating arrangement; b. Persons A and B must sit next to each other;

c. there are 4 men and 4 women and no 2 men or women can sit next to each other;

d. there are 5 men and they must sit next to each other;

3. A woman has 8 friends, of whom she will invite 5 to a party. How many choices does she have if 2 of the friends are feuding and will not attend together? How many choices does she have if 2 of hers friends will only attend together?

4. How many ways can a man divide 7 gifts among his 3 children if the eldest is to receive 3 gifts and the others 2 each

5. Suppose a precinct consists of 150 voters, 100 of whom are women and the remaining 50 are men. Suppose a sample of 25 voters will be selected in this precinct, how many possible…

a. ordered sample with replacement are there? b. ordered sample without replacement are there? c. unordered sample without replacement are there?

d. ordered sample with replacement are there when 10 of the voters selected are men?

e. ordered sample without replacement are there when 10 of the voters selected are men?

f. ordered sample without replacement are there when at least 20 of the voters selected are men?

Theorem (Complement Events): If A is an event then P(AC) = 1 – P(A)

 Thus, the sum of complementary events is 1; P(A) +P(Ac_{) = 1}

Theorem (Additive Rule): If A and B are any two events, then;  P(AB)= P(A)+P(B)- P(AB)

 Corollary: If A and B are mutually exclusive, then P(AB)= P(A)+P(B)

 Corollary: If A1, A2, A3, ... , An are mutually exclusive, then

 P(A1 A2A3 … An )= P(A1) + P(A2) + P(A3) + … +P(An)

Theorem: If A and Bare any two events, then using the previous two theorems;  P(ABC)= P(A) - P(AB)

Example: A health worker is studying the prevalence of certain diseases in a particular

community. Based on previous studies, the health worker was able to come up with the following figures: 10% of the people in the community will contract disease A sometime during their lifetime; 25% will contract disease B; and 5% will contract both diseases. Find the probability that a randomly selected person from this community will contract:

a) at least one of the 2 diseases b) disease B but not disease A c) exactly one of the 2 diseases

Solution:

Let A = event that selected person will contract disease A B = event that selected person will contract disease B

We can express the given percentages in terms of probabilities as follows:

P(A) = 0.10 P(B) = 0.25 P(AB) = 0.05

a) at least one of the 2 diseases; P(AB)= P(A)+P(B)- P(AB) = 0.10 + 0.25 - 0.05 = 0.30 b) disease B but not disease A; P(BAC)= P(B) - P(AB) = 0.25 -0.05 = 0.20

c) exactly one of the 2 diseases; P[(ABC)  (BAC) = P(ABC) + P(BAC)

= [P(A) - P(AB)] + [P(B) - P(AB)] = 0.05 + 0.20 = 0.25

6.5 Probabilities of Events

Theorem: If A and B are any two events where P(B>0) then; P(A|B) = ( ∩ ) _{( )}

There are times when we change our assignment of the probability of an event whenever we have additional information concerning the occurrence of other events.

The original measure of the probability without using additional information concerning the occurrence of other events is called an unconditional probability. While the probability measure derived using the information concerning the occurrence of other events that has already happened is called a conditional probability.

To reiterate the Conditional probability is the probability of event A occurring when we already know that some event B has already occurred.

Example: There are 100 insurance claims that are classified according to the type of policy and

whether the claim is fraudulent or not. If a claim is selected at random,

Category Type of Policy Total

Fire Auto Other

Fraudulent 6 1 3 10

Non-fraudulent 14 29 47 90

Total 20 30 50 100

Find the probability of:

a) selecting a fraudulent claim given that such claim is for a fire policy

b) selecting a fraudulent claim given that such claim is for a policy that is not about fire

Solution:

Let A = event of selecting a fraudulent claim B = event of selecting a fire policy

We can express the given data in terms of probabilities as follows:

P(A) = 10/100 = 0.10 P(B) = 20/100 = 0.20 P(AB) = 6/100 = 0.06 P(BC) = 80/100 = 0.80 P(ABC) = 4/100 = 0.04 a) P(A|B) = ( _{( )} ) ⁄ ⁄

6.5.1 Conditional Probability

Exercise:

1. A movie critic feels that the probabilities that a certain movie will get an award for best actress is 0.18, for best actor is 0.33, and at least one of these two awards is 0.40. Suppose it was just announced that the movie won the best actor award, what is the probability that it will win the best actress award?

2. The HR Department conducted a census to determine whether fear of flying is a major problem in their company. The employees were first classified as flyers (flown at least once), non-flyers likely to fly, or non-flyers not likely to fly. Then the employees were asked whether they get anxious of flying. The results of the census were as follows:

Anxiety Level Flight Experience Flyers Non-flyers Likely to Fly Non-flyers Not likely to Fly No anxiety 750 120 95 A little anxious 175 45 5 Very anxious 120 45 80

Two events A and B are said to be independent events if and only if any one of the following conditions is satisfied:

a) P(A|B) = P(A) if P(B) > 0; or b) P(B|A) = P(B) if P(A) > 0; or c) P(AB) = P(A)* P(B)

Otherwise, the events are dependent. Exercise:

1) The probability that Renzo will correctly answer the toughest question in an exam is 1/4. The probability that Sandro will correctly answer the same question is 4/5.

Assuming that the two events are independent (would not cheat), find the probability of the following events:

a. Event that both Renzo and Sandro will answer the question correctly b. Event that only Sandro will answer the question correctly

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