Baye's Theorem
In the previous section, we have learnt that
i) If A and B are two mutually exclusive events, then
ii)
Before we state and prove Baye's Theorem, we use the above two rules to state
the law of total probability. The law of total probability is useful in proving
Baye's theorem and in solving probability problems. Following is an example
which explains this law.
Example: Let S is the sample space which is the population of adults
in a small town who have completed the requirement for a college degree. The
population is categorized according to sex and employment status as follows
\ One of these individual is to be selected for a
tour throughout the country. Knowing that the individual chosen is employed,
what is the probability that the individual is a man?
Suggested answer: Let M be the event that a man is selected, E be
the event that the individual selected, is employed. Using the reduced sample
space, we have
Also, we have
From the original sample space, we have
Suppose that we are now given the additional information that 36 of those
employed and 12 of these unemployed are the members of the rotary club. What is
the probability of the event A that the selected individual is a member of the
rotary club?
Suggested answer: A is the event that the selected individual is a
member of the rotary club.
From the figure, it is clear that
The generalization of the foregoing example, where the sample space is
partitioned into n subsets is known as Law of Total Probability. Theorem: (Law
of Total Probability) If B1, B2, B3, �.., Bn
are mutually exclusive and exhaustive events of the sample space S, then for any
event A of S.
Proof:
From the venn diagram, we have
where
are mutually exclusive events.
Baye's Theorem Let S be a sample space.
If A1, A, A3 ... An are mutually exclusive
and exhaustive events such that P(Ai) ≠
0 for all i.
Then for any event A which is a subset of
We have,
Proof:
P(A)
P(A1) P(A/A1) +P(A2)P(A/A2)+�.+P(An)
P(A/An) �.(1)
Also,
Example: In a bolt factory 25%, 35% and 40% of the total is
manufactured by machines A, B and C, out of which 5%, 4% and 2% are respectively
defective. If the bolt drawn is found to be defective, what is the probability
it is manufactured by the machine A?
Suggested answer:
Given P (A) = 0.25, P (B) = 0.35 and P (C) = 0.4 Let D be the event of
getting a defective bolt.
P (D/A) = 0.05 P (D/B) = 0.04 P (D/C) = 0.02
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