Chapter six Elementary Probability1.1. Definition of basic terms of probability

Random experiment: - is a process of measurement or observation which is repeated at any time and whose out come cant be predicted with certainty. E.g. tossing a coin Outcome: - a particular result of an experiment (result of single trial of an experiment)Sample space: - is the set of all possible outcomes of a random experiment. Each possible outcome is called sample point.Event: - is a subset of a sample space (one or more outcomes of an experiment)Example1: if we toss a coin, the sample space (S) of this experiment is S = {head, tail} where head and tail are two faces of a coin. If we are interested the outcome of head will turn up then the event E= {head}Example 2: the sample space of tossing a coin twice is S= {HH, HT, TH, TT} Elementary or simple event: - an event has only one sample point.Mutually exclusive event: - two events E1 and E2 are said to be mutually exclusive if there is no sample point which is common to E1 and E2.

E1n E2 = i.e., if E1 and E2 are mutually exclusive events, then P (E1E2) = P (E1) + P (E2). Independent event: two events E1 and E2 are said to be independent if the occurrence or non occurrence of one cannot affect the occurrence or non occurrence of the other.Equally likely outcomes: - if each outcome in a sample space has the same chance to be occurred. Exa0mple In throwing a fair die all possible outcomes are equally likely. That means the elements of the sample space have equal chance to be occurred. Definition of probability Probability:-is a chance (likely hood) of occurrence of an event. It is expressed by a numerical value between 0 and 1 inclusively. Probability is a building block of inferential statistics. Counting techniques:In order to determine the number of out comes one can use several rules of counting.

1. Multiplication rule: - in a sequence of n events in which the first event has k1 possibilities, the second event has possibilities,, the nth event has kn possibilities, then the total possibilities of the sequence will be k1.k2.kn. Example: - in a personnel department a larger corporation wishes to issue each employee an ID cards with two letters followed by two digit numbers. How many possible ID cards can be imposed?Solution K1K2K3K4

26261010

Thus the total number of ID cards issued could be: 26*26*10*10=67600(with repetition)26*25*10*9=58500 (with out repetition) 2. Permutation: is an arrangement of n objects in a specific order. In this case order is crucial. a) The number of permutations of n objects taken all together is n! i.e. n! / (n-n)! b) The arrangement of n distinct objects in a specific order taking r objects at a time is given by nPr =n!/(n-r)!= n(n-1)(n-2)..(n-r-1)c) The number of permutation of n objects in which k1 are alike, K2 are alike, kn are alike is n! / k1!k2!....kn! Example: a photographer wants to arrange 3 persons in a raw for photograph. How many different types of photographs are possible? Solution: Assume 3 persons Aster (A), lemma (L), Yared (Y) and n=3 Since n! =3! = 3*2! = 6, there are 6 possible arrangement ALY, AYL, LAY, LYA, YLA and YAL Example2: fifteen athletes including Haile were entered to the race. a) In how many different ways could prizes for the first, the second and the third place be awarded? b) How many of the above triplets just counted have if Haile is in the first position? Solution: a) 15 objects taken 3 at a time 15P3=15! / (15-3)! = 2730 b) There are 14P2= 14! / (14-2) = 182

3. Combination: - counting technique in which the order of the objects is immaterial. Selection of r objects from a collection of n objects where r P (B) = P (AB)Rule 3: Suppose A and B are two events of a sample space, then

P (AB) = P (A) + P (B) - P (AB) Example: A fair die is thrown twice. Calculate the probability that the sum of spots on the face of the die that turn up is divisible by 2 or 3.Solution: 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),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} This sample space has 6*6 =36 elements let E1 be the event that the sum of the spots on the die is divisible by 2 and E2 be the event that the sum of the spots on the die is divisible by 3, then

P (E1 or E2) = P (E1 E2)

= P (E1) +P (E2) P (E1 E2) = 18/36 + 12/36 -6/36 = 24/36 = 2/3 Conditional probability: the conditional probability of an event A in relation to B is defined as the probability that event E occurs given that event A has already occurred.

P (A/B) = P (AB)/ P (B) where P (B) > 0

Remark: (i) P (AB) & P (B) are computed w. r. t. Original sample

(ii) P (S/B) = P(S B)/P (B) = P (B)/P (B) = 1

P (B/S) = P (B) because P (B/S) = P (BS)/P(S) = P (B)/1 =P (B)

(iv) if A and B are independent events, then P(A/B) =P(A) and P(B/A) =P(B) two events are independent if the occurrence of B doesnt affect the occurrence of A. i.e. P(A/B) =P(AB)/P(B)

P (AB) = P (A/B) *P (B) but P (A/B) = P (A)

Hence P (AB) = P (A)* P (B) Example: Suppose that an office has 100 calculating machines. Some of them use electric power (E) while others are manual (M) and some machines are well known (N) while others are used (U). The table below gives numbers of machines in each category. A person enter the office picks a machine at random and discovers that it is new. What is the probability that it is used with electric power?

EMTotal

N403070

U201030

Total 6040100

Solution: P (E/N) =P (EN) /P (N) = 40/70 =4/7

Chapter seven Probability Distributions

Probability distribution: is a list of all the possible out comes of an experiment and the probability associated with each out come. Example: Suppose we are interested in the number of heads showing face up on 3 tosses of coin. This is the experiment and the possible outcomes are 0 heads, 1 head, 2 head, and 3 heads. What is the probability distribution for the number of heads? Solution: The experiment has 8 possible outcomes, and below is the list of all the outcomes. Possible result Coin tossNo. of heads

1st 2nd 3rd

1.TTT0

2.TTH1

3.THT1

4.THH2

5.HTT1

6.HTH2

7.HHT2

8.HHH3

From the above table, the probability distribution for the number of heads isNo. of heads, xP (outcome), P (x)

01/8

13/8

23/8

31/8

Total1

6.1. Random variables. A random variable is a quantity resulting from an experiment that can assume different values.In any experiment of chance, the outcomes occur randomly. For example, rolling a single die is an experiment; and any one of the six possible outcomes can occur at a time.A random variable may be either discrete or continuous. i. Discrete random variable: a variable that results from counting and can assume only certain clearly separated values of some item of interest. Example: The number of heads in flipping a fair coin 5 times. ii. Continuous random variable: a variable that results from measuring and can take any value with in a certain range of values. Example: The distance b/n Sodo & Addis Ababa could be 330 km, 330.5 km, 331.5 km. and soon; depending on the accuracy of our measuring device. 6.2. Discrete probability distributions (probability mass function), expectation and variance of discrete random variableIf we organize a set of discrete random variables in a probability distribution, the distribution is called a discrete probability distribution; it is also called probability mass function (pmf). And it can be summarized by its mean and variance. Mean: The mean of a probability distribution is also referred to as expected value, E (x), and is given byMean = E (x) =(x p(x))P(x)= p (the possible value of random variable x). Variance & standard deviation: Though the mean is a typical value used to summarize a discrete probability distribution, it does not describe about the spread in the distribution, but the variance does this. 2= = 2 = x2p(x) 2

Standard deviation () =

Example: the following is the probability distribution for the number of cars a company expects to sell on a particular day. No. of cars sold, xProbability. P(x)

00.1

10.2

20.3

30.3

40.1

Total 1.0

1. What type of distribution is it? 2. On a typical day, how many cars does the company expect to sell? 3. What is the variance of the distribution? What is the standard deviation? Solution: 1. It is a discrete probability distribution. 2. = E (x) =(x p(x)) = 0(0.1) +1(0.2) +2(0.3) +3(0.3) +4(0.1) = 2.1. Interpretation: Over a large number of days, the company expects to sell 2.1cars a day. Of course, it is not possible for him to sell exactly 2.1 cars on any particular day; thus the mean is sometimes called the expected value. 3. 2 = x2p(x) 2 = (02(0.1)+12(0.2)++42(0.1)) - (2.1)2 = 1.29

==

6.3. Common discrete problem distributions 1. Binomial distribution. It is used to represent the probability distribution of discrete random variables. Binomial means two categories. The successive repetition of an observation (trial) may result in an outcome which possesses or which does not possess a specified character. Our primary interest will be either of these possibilities. Conventionally, the outcome of primary interest is termed as success. The alternative outcome is termed as failure. These terminologies are used irrespective of the nature of the outcome. For example, non-germination of a seed may be termed as success.Properties: 1. There must be only two mutually exclusive outcomes: success or failure. 2. The probability of success, p, and the probability of failure, q=1-p, remains constant from one trial to another. 3. The probability of success in one trial is totally independent of any other trial. 4. The experiment can be repeated many times Example: The coin flip experiment has only two possible outcomes: head or tail. The probability of each is known and constant from one trial to another. We can flip a coin many times. The binomial distribution is computed by

C = combination n= number of trialsx=number of successesp=the probability of successq=1-p=the probability of failure Mean of a binomial distribution = npVariance of a binomial distribution 2 = npqExample: There are 5 flights daily from Addis Ababa to Washington, suppose the probability that any flight arrives late is 0.2. What is the probability that a. None of the flights are late today?b. Exactly one flight is late today?c. Construct the entire probability distributiond. What is the probability that more than 4 flights are late?e. Between 2 and 4 (inclusive) flights are late?f. What is the mean?g. What is the variance?Solution: given that the probability of a particular flight is late is 0.2, and thus the probability that a particular flight is not late is 0.8. There are 5 flights, so n = 5, and x refers to the number of successes. In the questions a to e, we are asked about the late flights, so here let success = late flight. Then p = 0.2, and q = 0.8. a. P (none of the flights are late today) = P (0 flights are late) = P (x = 0)

=0.3277b. P (exactly one flight is late today) = P (1 flight is late) = P (x = 1)

c. The entire distribution is Number of late flights, xP (x)

00.3277

10.4096

20.2048

30.0512

40.0064

50.0003

Total 1.0000

d. P (x > 4) = P (x = 5) = 0.0003e. P (2 x 4) = P (x = 2) + P (x = 3) + P (x = 4) = 0.2048 + 0.0512 + 0.0064 = 0.2624f. = np = 5 * 0.2 = 1 late flight or 5 * 0.8 = 4 not late flightsg. 2 = npq = 5 * 0.2 * 0.8 = 0.82. The Poisson distribution The Poisson distribution is also used to represent the probability distribution of a discrete random variable. It is employed in describing random events that occur rarely over some unit of time or space. Examples of events where Poisson probability function can be used: Number of telephone calls per hour Number of typing errors per page Number of accidents on a particular road per day Hospital emergencies per day, etcAssumptions: 1. The probability of occurrence of an event is constant for any two intervals of time or space2. The occurrence of an event in any interval is independent of the occurrence in any other interval.Having these assumptions, the Poisson distribution is given by the function

P (x) = Where x = the number of times the event has occurred

= is the mean no. of occurrences per unit of time or space. e = 2.71828, the base of the natural logarithm system. Example: Simple observation over the past 80 hours has shown that 800 customers have entered the shop. What is the probability that a. exactly 5 customers will arrive during any given hour?b. more than 3 customers will arrive during any given hour?c. exactly 5 customers will arrive during any 30 minutes?

Solution: = a. P (x = 5) = b. P (x > 3) = P (4) + P (5) + by the complement rule that we have discussed earlier P (x > 3) = 1 P (x 3)

= = 1 - = 1 (0.0103) = 0.989c. P (x = 5/30 minutes) Here, as we are asked per 30 minutes, we should change the value per 30 minutes; thus

=

P (x = 5) =

6.4. Continuous probability distributionContinuous probability distribution is also called probability density function (pdf)Let x be a continuous random variable, then the pdf of x is a function f(x), such that for any two numbers a and b with a b

P (a) = Which is the area under the curve bounded by x=a and x=b

If f(x) is pdf of x 1. f(x)0 for all x2. i.e. area under the graph of f(x) must equals 1, since the sum of relative frequencies is 1. Example: The diameter of an electronic cable, say x, is assumed to be continuous random variable with pdf f(x)=6x(1-x), 01. Check f(x) is pdf 2. Determine number b such that P(0.5