Chapter 3: Business Processes and Knowledge Management
3.7 Complexity and Knowledge Management
3.3.1 Additive Law
It is also known as “either” “or” rule. The probability that either A or B will occur mutually is the sum of their independent probabilities.
For example, two coins are tossed. The probability of either two heads or tails will occur is
P(TT + HH)
Therefore, P(A + B) = P(A) + P(B) − P(AB) for joint occurrence of events A and B at the same time
For mutually exclusive events, A and B;
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) This can further be explained by use of venn diagram
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Example
What is the probability that in a mendelian cross between goats that are heterozygote for color Bb will produce offspring having heterozygote?
Solution
There are two ways in which a zygote may be produced
1. The dominant zygote allele (B) may be in the egg and the recessive (b) in the sperm or
2. The dominant allele may be in the sperm and the recessive in the egg.
Therefore the probability that the offspring will be heterozygous is the sum of the probabilities of those possible ways.
a. The probability that the dominant allele will be in the egg with recessive in the sperm is;
1 2 ×
1 2 =
1
b. The probability that the dominant allele will be in the sperm and the 4 recessive in the egg is
1 2 ×
1 2 =
1
Therefore the probability that heterozygous will be produced is 4 1
4 × 1 4 =
2 4 =
1 2
3.3.2 Multiplicative Rule of Probability
The multiplicative rule suggests that for event A or B that are independent.
They probability that they will both occur is equal to the product of their individual probabilities.
P (A and B) = P (A ∩ B) = P(A) × P(B) This is also referred to as the “Both” “And” rule.
Permutation
It is the number of ways the individual elements of interest are arranged in order. Those elements appear once. ∩ ! Denotes the product of all number of elements ranging from 1.∩
Pn = n! = (1,2, 3 …n) Example; 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
61 If we want to select only few objects in n group then permutation will be the best for it.
nPr = n!
(n − r)!
Example; if we have 10 cows on our farm, how many different ways can 5 cows be selected from the farm.
Solution: Using
nPr = n!
(n − r)!
Our n = 10 cows
r = 5 possible ways hence
,P
= 10!
(10 − 5)! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 5 × 4 × 3 × 2 × 1
= 3,628,800
120 = 30, 240 Combination
Just like the permutation, combination involves the number of ways individual members of the n group are arranged but in combination. The order by which they are arranged is not important/ considered. C =
! (.)!
That is the combination of n taken at a time
Example; If 7 cows are taken from a herd of 20, how many cows combination exists.
C = 20
(20 − 7)7! = 20 × 19 × 18 … 1 13 × 7 × 6 × 5 … 1
3.3.3 Compound Events
These are events that have more than one event. Assuming A and B occur at the same time. A and B could be said to have an intersection of the event(A ∩ B)
Just like the addition rule, an instance where “either” A “or” B occur is called union of events (A ∪ B)
The probability of A ∩ B = P(A ∩ B) The probability of A ∪ B = P(A ∪ B)
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The probability of an event that A does not occur (A complement) denoted by Ac is
P(A%) = 1 − P(A) Example
Let the event B be such that the result of a throw of die is an even number, let event A be such that the number is less than 4
B = 2,4, 6 Zind B ∩ A A = 1, 2, 3ZindP(A ∩ B) Solution
B ∩ A = 2
P (A ∩ B) = P(A) + P(B) = 1 6
3.3.4 Conditional Probability
Is the probability of an event B happening when one have a prior knowledge of the probability that some other event A has occurred conditional or statistical dependence exist when the probability of an event is dependent upon the occurrence of some other event.
P B A = P(A ∩ B) P(A)
For independent probability, the occurrence of event A is not affected by the occurrence of B. Therefore the probability that both occur is the product of their individual probability
P(A ∩ B) = P(A) P(B)
But for dependent event, the probability of B occurring is affected by the occurrence of A
P B A = P(A ∩ B) P(A) Example of independent and dependent probability
1. If we throw a coin and A and B are the possible occurrences, then P(A ∩ B) = P(A)P(B) = 1 2 × 1 2 = 1 4
2. For dependent probability: if in a pen of 40 rabbits, divided into 4 colors (black 10, grey 10, white 10, and brown 10) if the first draw is black and the second is black again, then P (A ∩ B).
P(A ∩ B) = P(A)P B A
= If black is drawn and the second draw is black.
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= 10 40 9
39 = 90 1560 The probability of drawing two black rabbits =
,3
3.3.5 Bayes’ Theorem
The theorem is useful in evaluating the probability of an event A with provided information on the probability of another event B that had occurred after the occurrence of A.
PB A =
P(B) P(A B )
(P(B)P A B + (P(B)P(A B) ) P(B) = probability of event B compliment P A B
= probability of event A having known that the complement B has occured Example
If two strains of cock A and B are used in a flock of chickens, if cock A is used, 80% of the hens and cock B, 20% of the hens. We assume that recorded fertility of eggs fertilized by cock A and B are 45% and 95%
respectively. For a certain egg, the information about the cock is missing, what is the probability that the cock has fertilized the egg is cock B.
P(A,) = 0.8 is the probability of using cock A P(B,) = 0.2 is the probability of using cock B E = The event that an egg was layed from a cock P E A ,
= 0.45 The probability of successful making or fertility of cock A P E B
= 0.95 The probability of a successful making or fertility of cock B Therefore the probability that the egg is fertilized by the cock is B
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P B = E P(B ∩ E) P(E) =
P(B) P(E B )
P(A)P(E A ,)+ P(B) P(E B )
= (0.2) (0.95)
(0.8)(0.45) + (0.2)(0.45) = 0.19
1.25 + 0.09 = 0.142 The probability that the cock that fertilize the egg is cock B = 0.142 SELF-ASSESSMENT EXERCISE
Name the types of probability.
4.0 CONCLUSION
Looking at how probabilities help us in predicting the outcome of some events even before they happen. You must have seen how we can apply probability to most of our day to day lives and our experiment to help us predict the outcome of some events.
5.0 SUMMARY
In this unit, you learnt the meaning of probability, the basic terms used in probability, the various types of probability as well as various probability rules. These are tools we use for predicting the outcome of an occurrence so that we will prepare for them.