Probability
Objectives
By the end of this article, you should be able to:
- Describe Probability Theory.
- Describe Theoretical and Empirical Probability.
- Solve probability problems using rules of probability
- Solve probability problems using Bayes Theorem
Probability Theory
Probability is a concept of quantified chance, likelihood, possibility, or proportion. It can be considered as a measure of likelihood of an event occurring. For example, Assume you have 3 buckets (A, B, & C) of sand with nails having different colours. Bucket A has 2 red nails and 1 green nail. Bucket B has 3 red nails and 1 green nail. Bucket C has 4 red nails and 2 green nails. If you pick a green nail one wins some prize. From this information we see that Bucket A has better chances of winning a prize than bucket B. Bucket A & C are the best options because proportion of green nails is highest. Other examples of Probability Statements are:
- There is a 30% chance that this job will not be finished in time.
- There is every likelihood that the business will make profit next year.
- Nine times out of ten he arrives late for his appointments.
- There is no possibility of delivering the goods before Tuesday.
We can also have a probability problem like “Emmanuel Radio Station broadcasts genetically modified awareness issues for 15 minutes on the hour, every hour. Farmers in a certain district listen to this radio station at random times when they feel like. Find the probability that farmer X, who switches on his/her radio, tuned to the radio station finds genetically modified issues being broadcast”.
Probability theory is quite closely linked to Set theory. Set theory talks about the sample space, subset, and events. Sample space is a set of ALL possible outcomes of a situation.
For example, If a die is tossed once, the sample space S is {1, 2, 3, 4, 5, 6}. Subset is a collection of some members of sample space. For example, if we let S = {1,2,3,4,5,6} be the sample space, then A= {2,4,6} is a sub set. Event is a subset of a sample space. For example, if we let S = {1,2,3,4,5,6} be the sample space then B={1,3,5} is an event of getting an odd number.
Events can be mutually exclusive or independent. In terms of mutually exclusive events, two events of the same experiment are mutually exclusive if their respective event sets do not overlap i.e. when the experiment is performed mutually exclusive events cannot happen at the same time (cannot occur together). For example, if a die is tossed, event B={1,3,5} (event of getting an odd number) and event A={2,4,6} (event of getting even number) are mutually exclusive. However, events “Even number” and “A number less than 4” are NOT mutually exclusive.
For independent events, two events are independent if the occurrence (or not) of one of the events will in no way affect the occurrence (or not) of the other i.e. the two events are defined on two physically different experiments. For example, events “ a particular supplier will deliver on time” and “a particular customer will not be paid within a specified time” are independent events because they are each based on physically different situations or experiments. Please Note that Two mutually exclusive events CAN NOT be independent since the occurrence of one excludes the other (i.e. they are dependent in a particular way).
Theoretical Probability
Theoretical probability is the name given to probability that is calculated without the experiment being performed—It is calculated using only information that is known about the physical situation. If E is some event of an experiment that has an equally likely outcome set U, then the theoretical probability of event E occurring when the experiment is performed is written as P(E) and given by:
P(E) = number of different ways that the event can occur/number of different outcomes of the experiment i.e. n(E)/n(U) where: n(E) is number of outcomes in event set E; n(U) is the total possible number of outcomes (in outcome set U)
Note: We write Pr(A) or P(A) or pr (A) to mean probability that we get event A.
Example: What is the probability of getting an even number when one tosses an unbiased (fair) die?
Solution
- Let S = {1,2,3,4,5,6}
- Let A = {2,4,6} be event of getting even number
- Pr(A) = n(A)/n(S) where n is the number of members in A & S
- Pr(A) = 3/6
- Pr(A) = 1/2
Example: A wholesale stationer stocks heavy (2B), medium (HB), fine (2H) and extra fine (3H) pencils which come in packs of 10. Currently in stock are 2 packs of 3H, 14 packs of 2H, 35 packs of HB and 8 packs of 2B. If a pack is chosen randomly for inspection, what is the probability that they are medium (HB)?
Solution
The probability of choosing a medium pencil pack is given by the number of medium pencil packs divided by the total number of pencil packs.
- That is, P (medium) = n(medium)/n(U)
- P(medium) = 35/ (2+14+35+8)
- P(medium) = 35/59
- P(medium) = 0.59
Empirical Probability
Empirical probability is the name given to probability that is calculated using the results of an experiment that has been performed a number of times. It is sometimes called relative frequency or subjective probability. It is used in situations where an outcome set for an experiment is not known OR if the known outcomes are not equally likely. If E is an event of an experiment that has been performed a relatively large number of times to give a frequency distribution then the empirical probability of event E occurring when the experiment is performed one more time is given by:
P(E) = number of times the event occurred/number of times the experiment was performed.
- P(E) = f(E)/∑ f Where: f(E) is the number of times that event E has occurred
- ∑ f is the total frequency = number of performances of experiments
The empirical probability is simply the proportion of times that event E actually occurred when the experiment was performed. For example, If out of 60 orders received so far this financial year, 12 were not completely satisfied, the proportion 12/60 = 0.2 is the empirical probability that the next order received will not be completely satisfied.
Rules of Probability
Rule 1: Probability Limits. The probability of any event A occurring must lie between 0 and 1 inclusive. If we let A be an event, then 0≤ Pr(A) ≤1
Example: If one tosses a fair die, what is the probability that you get a 1,2,3,4,5 or 6?
Solution
- S = {1,2,3,4,5,6}, A = {1,2,3,4,5,6}
- P(A)= n(A)/n(S), 6/6, = 1
Probability Value of 1 is called Certainty—One is sure that such an event will occur without fail. Other certain events include an employee being either male or female.
Example: If one tosses a fair die, what is the probability of the outcome of the die being 7?
- S = {1,2,3,4,5,6}, A={ }—Empty set
- P(A)= n(A)/n(S), 0/6, = 0
Probability value of 0 means the event is an Impossibility. Other impossible events include an employee of a workforce who is less than 1 year old.
Rule 2: Total Probability. The sum of the probabilities of all possible outcomes of an experiment must total exactly 1. In the tossing of die example, P(1)+ P(2)+ P(3)+P(4)+P(5)+P(6) = 1
Rule 3: Complementary. Let A be an event, then the complement of A is denoted A- or Ac or Ā. And
P(A) = 1-P(Ac )
Example: If the probability that a fish will die with concentration of 10mg of mercury is ¼. Find the probability that the fish will NOT die with concentration of 10mg of mercury?
Solution:
Let A be event that the fish dies with 10mg of mercury.
P(Ac ) = 1-P(A),
1-1/4,
=3/4
Rule 4: Addition. Suppose a fair die is tossed once, find the probability of (i) A or B occurring (ii) A and B occurring if A = {2,4,6} and B ={1,2,5}
Solution
- S = {1,2,3,4,5,6}
- P(A or B) = n(A or B)/n(S), = 5/6
- P(A & B) = n(A&B)/n(S), = 1/6
“OR” is used inclusively—A member of A or B is in at least one of the two sets.
In general the Pr(A or B) = P(A) + P(B) – P(A&B)
Example: If the probability that it rains well next year is 0.5 and the probability that we will have high yields next year is 0.6. If the probability that it will rain well next year and we will have high yield is 0.2. Find the probability that it will either rain well next year or we will have high yield?
Solution
Let A be event of raining well next year and B event of high yield next year.
- Given P(A) = 0.5, P(B) = 0.6, P(A&B) = 0.2
- P(A or B) = 0.5+0.6-0.2, = 0.9
For Mutually Exclusive Events, If A and B are any two mutually exclusive events of an experiment then the formula or rule becomes:
Pr(A or B) = P(A) + P(B)
This is because mutually exclusive events do not overlap hence do not have P(A&B).
Rule 5: Independent Probability (Multiplication Rule)
Two events A and B are said to be independent if P(A/B) = P(A). For example, if we let B be the event of getting a number less than 3 when a fair die is tossed and A be the event of getting an even number on the die then:
- P(A) = 3/6 = 1/2; P(B) = 2/6 = 1/3; P(A&B)= 1/6
- P(A/B) = P(A&B)/P(B) = {1/6 }/{1/3}= 1/2
Therefore A and B are independents. Information about B was not necessary or needed at all as A does NOT depend on it.
In general, this means that P(A&B) = P(A)* P(B)
Rule 6: Conditional Probability. This is a probability of an event occurring given that another event has occurred. The two Events must interact in some way (i.e depend on each other).
Example: A fair die is tossed once. Find the probability that an even number (score) appears if the outcome is a score less than 6.
Solution
- Let S = {1,2,3,4,5,6}, A = {2,4,6} event of getting even number and B = {1,2,3,4,5} event of getting a score less than 6.
- Pr(A) = n(A)/n(S), = 3/6 = ½
- Pr(A/B) = 2/5—probability of A given B. We only need to concentrate on event B.
So P(A/B) = n(A & B)/n(B). Dividing the numerator and the denominator by n(S) gives:
In general Conditional Probability is calculated as P(A/B) = P(A&B)/P(B)
Example: In our rainfall yield example, we had P(A&B) = 0.2, P(B) = 0.6. Therefore P(rains next year/we will have high yield next year) will be:
P(A/B) = P(A&B)/P(B), =0.2/0.6, =1/3
Example: Of 8 equal candidates for a job, 3 are qualified accountants, 4 are graduates and 2 have neither of these qualifications. Find the:
a. probability that a graduate gets the job.
b. probability that he is a graduate given that a qualified accountant has got the job.
c. probability that a qualified accountant gets the job, given that a graduate did not get the job.
Solution
a. Pr (graduate) = 4/8; = 0.5
b. Out of the 3 qualified accountants, only 1 is a graduate. Thus, Pr(graduate/qualified accountant) = 1/3; = 0.33
c. 4 of the candidates are non-graduates, and of these, 2 are qualified accountants.
Thus, P(qualified accountant/non-graduate) = 2/4; = 0.5
Note: Venn diagrams may be employed to represent situations
Based on rule 5, for any events A and B, the general multiplication rule becomes Pr(A & B) = Pr(A) * Pr(B/A). Now note that If A and B are Independent events, Pr(B/A) = Pr(B) and the above rule reverts to the ordinary multiplication rule P(A&B) = P(A) * P(B)
A problem on conditional probability can also be solved by means of a Tree diagram. Let us look at this example: A man goes to work on foot, by bus or by car with respective probabilities of 0.5, 0.2 and 0.3 respectively. If he goes on foot, the probability that he arrives to work late is 0.4. If he goes by bus, the probability that he arrives to work late is 0.7 and if he goes by car, the probability that he arrives to work late is 0.5. Determine the probability that: he is late on a given day, he travelled by bus given that he is late.
Solution
- Let F = man goes on foot, B = man travels by bus, C = man travels by car and L = man is late.
- Then the Tree diagram becomes or will look like:
a) The probability of the man being late, irrespective of the means of transport used, is given by:
- P(L)= P(F)P(L/F)+ P(B)P(L/B)+ P(C)P(L/C)
- = (0.5)(0.4) + (0.2)(0.7) + (0.3)(0.5) = 0.49
b) The probability that the man travelled by bus given that he is late is given by:
- P(B/L) = [P(B)P(L/B)] / [P(F)P(L/F)+P(B)P(L/B)+P(C)P(L/C)]
- 0.14/0.49, = 2/7. This could mean contribution of the bus to his lateness.
Bayes Theorem
Bayes Theorem is a formula which can be thought of as “reversing” conditional probability. It finds a conditional probability (A/B) given, among other things, its inverse (B/A). Bayes Theorem is given by:
Pr(A/B) = {Pr(A)*Pr(B/A)}/Pr(B)
Example: If the probability of meeting a building contract date is 0.8, the probability of good weather is 0.5 and the probability of meeting the date given good weather is 0.9. Calculate the probability that there was good weather given that the contract date was met.
Solution
- Let G = Good weather and M = Contract date is met
- Given Pr(M) = 0.8; Pr(G) = 0.5 and Pr(M/G) = 0.9
- Therefore Pr(G/M) = {Pr(G)*Pr(M/G)}/Pr(M)
- Pr(G/M) = {0.5 * 0.9}/0.8 = 0.56
- Thus the probability that there was good weather given that the contract was met is 0.56