PROBABILITY
Definition: The probability of a given event is an expression of likelihood of occurrence of an event .A
Probability is a number which ranges from 0 to 1.
0 for an event which cannot occur
1 for an event which can occur
Formula :
𝐹𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙 𝑒 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑎𝑠𝑒𝑠
Importance for the concept of Probability
According to Ya-lun Chou:
“Statistics as a method of decision-making under uncertainty is founded on probability theory, since probability is at
once the language and the measure of uncertainty and the risks associated with it.”
Experiment: The term experiment refers to processes which result in different possible outcomes or observations.
Random Experiment:
1) all possible outcomes are known in advance.
2) None of the outcomes can be predicted with certainty.
Note:
Each performance of a random experiment is called a Trial and the result of a performance (or trial) is called an
outcome or case.
Sample Point:
The outcome of the random experiment is called a sample point.
Sample space:
The set of all possible outcomes (or sample points) of a random experiment is called a sample space.
Finite Sample space:
A sample space is said to be finite if it contains a finite number of sample points.
Infinite Sample space:
A sample space is said to be infinite if it contains a infinite number of sample points.
Discrete Sample space:
A sample space is said to be discrete if it contains finite or finitely many or countably infinite sample points
which can be arranged into a simple sequence E1,E2…….. Corresponding to natural number.
Continuous Sample space:
A sample space is said to be continuous if it contains non-denumerable (uncountable)
points.
number of sample
Events:
A subset of a sample space is called an event. An empty subset φ of S us called impossible event and the space
itself is called a certain event.
The subset containing a simple sample point is known as simple event or elementary event.
PROBABILITY OF AN EVENT:
The probability of any event is equal to the sum of the probabilities of the sample points in the event
Types of event
Equally like Events:
Two or more events are said to be equally likely if any one of them cannot be expected to
preference to the others.
Example:
In a tossing of an unbiased coin, Head is likely to come up as tail.
occur in
Composite Event:
The union of two events A & B denoted by A U B and A + B is called a composite event of A & B
Compound Event:
The intersection of two events A & B denoted by A∩B or AB is called compound event of A & B.
Mutually Exclusive (or incompatible) Events:
Two events are said to be mutually exclusive if they have no sample point in common i.e.
A∩B =Ф
Exhaustive events:
Two events are exhaustive if their union is equal to the sample space.
Independent Events:
Two or more events are said to be independent if the occurrence of one does not affect the occurrence of the
other.
Calculate probability in case of independent events:
1) Probability of occurrence both A & B i.e. P(A ∩ B) = P(A) X P(B)
2) Probability of Non-occurrence both A & B i.e. P(A ∩ B) = P(A) X P(B)
3) Probability of occurrence A & not B i.e. P(A ∩B) = P(A) X P(B)
4)
5)
6)
7)
Probability of occurrence B & not A i.e. P( Ā ∩ B) = P( Ā) X P(B)
Probability of occurrence at least one event i.e. P( A U B) = 1-[P(Ā) X P(B)]
Probability of non occurrence at least one event i.e. P(Ā UB) = 1 –P (A∩ B) or 1- [ P(A) X P(B)]
Probability of occurrence only one event P(A ∩ B) + P( Ā ∩ B) = [ P(A) X P(B) + P( Ā) X P(B) ]
Dependent Events:
Two events are said to be dependent if the occurrence of one affects the occurrence of the
other.
Calculate probability in case of dependent events
1)
2)
3)
4)
5)
Probability of occurrence both A & B i.e. P(A ∩ B) = P(A) + P(B) – P(AUB)
Probability of Non-occurrence both A & B i.e. P(A ∩ B) = 1-P(AUB)
Probability of occurrence A & not B i.e. P(A ∩ B) = P(A) - P(A∩B)
Probability of occurrence B & not A i.e. P( Ā ∩ B) = P( B) - P(A∩B)
Probability of non occurrence at least o A OR B i.e. P(Ā U B) = 1 –P (A∩ B)
Mutually exclusive & Exhaustive Events:
A number of events are said to be mutually exclusive & exhaustive events if
i) every two of them are mutually exclusive
ii) one of them necessarily occurs in any trial.
Experiment Counting Rules & Assignment Probabilities
we define an experiment as a process that generates well-defined outcomes On any single repetition of an experiment, one and only
one of the possible experimental outcomes will occur. Several examples of experiments and their associated outcomes follow
TREE DIAGRAM : A tree diagram is a graphical representation that helps in visualizing a multiple-step experiment
Addition Law: The addition law is helpful when we are interested in knowing the probability that at least one of two events occurs.
That is, with events A and B we are interested in knowing the probability that event A or event B or both occur
P(AUB) =P(A) + P(B) – P(A∩B)
UNION OF TWO EVENTS: The union of A and B is the event containing all sample points belonging to A or B
or both. The union is denoted by AU B
INTERSECTION OF TWO EVENTS
Given two events A and B, the intersection of A and B is the event containing the sample points belonging to both A and B. The
intersection is denoted by A∩B
CONDITIONAL PROBABILITY: Two events are said to be dependent when B can occur only when A is known to have
occurred( or vice versa). The probability associated with such an event is called the conditional probability
Question: A bag contains 5 white and 3 black balls. Two balls are drawn at random one after the another without
replacement. Find the probability that both balls are black
Solution:
Total number of balls = 5 + 3 = 8
Probability of drawing a black ball in first attempt =
𝟑
𝟖
Probability of drawing the second black ball given that the first ball drawn is black
P(B/A) =
𝟐
𝟕
𝟑
𝟐
𝟑
𝟖
𝟕
𝟐𝟖
Probability that both balls drawn black P(AB) = P(A) X P(B/A) = * =
Question: Suppose that we have two events, A and B, with P(A)=0.50, P(B)=0.60, and P(A∩ B)=0 .40.
a. Find P (A / B).
b. Find P (B / A).
c. Are A and B independent? Why or why not?
Solution:
Question :A purchasing agent placing rush orders for a particular raw material with two different suppliers A & B.If either
orders arrive in four days, the production process must be shut down until at least one of the order arrives. The probability
that supplier A can deliver the material in 4 days is 0.55. The probability that supplier B can deliever the material in four days
is 0.35
A) What is the probability that both suppliers will deliver the material in four days ? because two separate suppliers
are involved, we are willing to assume independence.
B) What is the probability that at least one supplier will deliver the material in 4 days
C) What is the probability that the production process will be shut down in 4 days because of a shortage of a
material
Solution :
Question: find the probability of drawing a queen, a king and a knave(jack) in that order from a pack of 52 cards in
three consecutive draws, the cards drawn not being replaced.
Solution : DO YOURSELF ANS : 0.00048
BAYE’S THEOREM
we begin the analysis with initial or prior probability estimates for specific events of interest. Then,from sources such as
a sample, a special report, or a product test, we obtain additional information about the events. Given this new
information, we update the prior probability values by calculating revised probabilities, referred to as posterior
probabilities. Bayes’ theorem provides a means for making these probability calculations
Example : Tree Diagram for Two supplier and P(A1) = 0.65 and P(A2) = 0.35
BINOMIAL DISTRIBUTION
Question: Consider a binomial experiment with two trials and p =0 .4.
a. Draw a tree diagram for this experiment (see Figure 5.3).
b. Compute the probability of one success, f (1).
c. Compute f (0).
d. Compute f (2).
e. Compute the probability of at least one success.
f. Compute the expected value, variance, and standard deviation
Solution :
a) Tree Diagram
b) Here n= 2 , p=0.4 and q =0.6 , r=1
Then f(1) = nCrPrqn-r = 2C1 (0.4)1(0.6)1 = 0.48
c) Here r=0
f(0) = 2C0 (0.4)0(0.6)2 = 0.36
d) Here r=2
f(2) =2C2 (0.4)2(0.6)0 = 0.16
e) Here (r≥ 1) = f(1) + f(2 ) = 0.48+0.16 =0.64
f) Expected value = np = 2*0.4= 0.8
Variance = npq = 2*.4*.6=0.48 ,
Standard deviation σ = √0.48 = 0.6928
Questions: When a new machine is functioning properly, only 3% of the items produced are defective. Assume that
we will randomly select two parts produced on the machine and that we are interested in the number of
defective parts found.
a. Describe the conditions under which this situation would be a binomial experiment.
b. Draw a1 tree diagram similar to Figure 5.3 showing this problem as a two-trial experiment.
c. How many experimental outcomes result in exactly one defect being found?
d. Compute the probabilities associated with finding no defects, exactly one defect, and two defects
Solution:
Question: Military radar and missile detection systems are designed to warn a country of an enemy attack. A reliability question is
whether a detection system will be able to identify an attack and issue a warning. Assume that a particular detection system has
a .90 probability of detecting a missile attack. Use the binomial probability distribution to answer the following questions.
a. What is the probability that a single detection system will detect an attack?
b. If two detection systems are installed in the same area and operate independently, what is the probability that at
least one of the systems will detect the attack?
c. If three systems are installed, what is the probability that at least one of the systems will detect the attack?
d. Would you recommend that multiple detection systems be used? Explain.
Solution :
a) Probability that a single detection system will detect an attack = 0.90
Question: Fifty percent of Americans believed the country was in a recession, even though technically the economy had not
shown two straight quarters of negative growth (BusinessWeek, July 30, 2001). For a sample of 20 Americans, make the
following calculations.
a. Compute the probability that exactly 12 people believed the country was in a recession.
b. Compute the probability that no more than five people believed the country was in a recession.
c. How many people would you expect to say the country was in a recession?
d. Compute the variance and standard deviation of the number of people who believed the country was in a recession.
Solution :
Question : A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that
20 students registered for the course.
a. Compute the probability that two or fewer will withdraw.
b. Compute the probability that exactly four will withdraw.
c. Compute the probability that more than three will withdraw.
d. Compute the expected number of withdrawals.
Solution :
POISSONS DISTRIBUTION
Question: Consider a Poisson distribution with μ= 3.
a. Write the appropriate Poisson probability function.
b. Compute f (2).
c. Compute f (1).
d. Compute P(x= 2).
Solution :
Question : During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two
minutes.
a. What is the expected number of calls in one hour?
b. What is the probability of three calls in five minutes?
c. What is the probability of no calls in a five-minute period?
Solution :
Question : The national safety council reported that air bag-related fatalaties dropped to 18 in the year 2000
A) Compute the expected number of air bag-related fatalities per month
B) Compute the probability of no air bag-related air fatalities in a month
C) Compute the probability of two or more air bag related fatalities in a month.
Solution : A) Expected number or average or mean of air bag-related fatalities per month = 18/12 = 1.5
B) no air bag it means r=0